In mathematics, a sequence is an ordered list of objects (or events). It contains members (also called elements or terms), and the number of terms (possibly infinite) is called the length of the sequence. Unlike a set, order matters, and exactly the same elements can appear multiple times at different positions in the sequence. A sequence is a discrete function.
Key Characteristics
Order Matters: \((1,2)\) is a different sequence from \((2,1)\).
Repetition Allowed: ( \(0,0,1\) ) is a valid sequence, distinct from \(\{0,1\}\).
Length: Can be finite (e.g., (1, 2, 3)) or infinite (e.g., (1, 2, 3, …)).
Discrete Function: Maps positions (1st, 2nd, 3rd…) to terms, forming a rule.
Examples
Finite and Infinite Sequences
On the basis of the number of terms, there are two types of sequences.
Example 1: Consider the sequence defined by \(a_n=a n^2 +b n+c\). If \(a_1=1, a_2=5\) and \(a_3=11\) then find the value of \(a_{10}\).
Solution: \(a_1=1 \Rightarrow a+b+c=1 \dots(i)\)
\(
\begin{aligned}
& a_2=5 \Rightarrow 4 a+2 b+c=5 \dots(ii) \\
& a_3=11, \Rightarrow 9 a+3 b+c=11 \dots(iii)
\end{aligned}
\)
Now from (ii) – (i), we have \(3 a+b=4 \dots(iv)\)
From (iii) – (ii), we have \(5 a+b=6 \dots(v)\)
From (v) – (iv), we have \(2 a=2\) or \(a=1\)
\(\Rightarrow b=1\) (from (iv)), and \(c=-1\) (from (i))
Hence \(a_n=n^2+n-1\)
Hence \(a_{10}=100+10-1=109\)
Series
By adding or subtracting the terms of a sequence, we get an expression which is called a series. If \(a_1, a_2, a_3, \ldots, a_n\) is a sequence, then the expression \(a_1+a_2+a_3+\cdots+a_n\) is a series. For example,
(i) \(1+2+3+4+\cdots+n\)
(ii) \(2+4+8+16+\cdots\)
Progression
It is not necessary that the terms of a sequence always follow a certain pattern or they are described by some explicit formula for the \(n\)th term. Those sequences whose terms follow certain patterns are called progressions.
Arithmetic Progression (A.P.)
Defintion: \(A\) sequence \(a_1, a_2, a_3, \ldots, a_n, \ldots\). is called an arithmetic progression (A.P.), if the difference of any term and the previous term is always same.
i.e., \(\quad a_{n+1}-a_n=\) Constant \((=d)\) for all \(n \in N\)
or, \(a_{n+1}-a_n\) is independent of \(n\).
The constant difference ‘ \(d\) ‘ is called the common difference.
General term (\(n^{\text {th }} \text { term from the beginning }\)): The \(n^{\text {th }}\) term \(a_n\) of an A.P. with first term ‘ \(a\) ‘ and common difference ‘ \(d\) ‘ is given by
\(
a_n=a+(n-1) d= l \text { (last term), where } d=a_n-a_{n-1}
\)
If an A.P. consists of finite number of terms, say, \(m\). Then, \(n\)th term from the end is \((m-n+1)^{\text {th }}\) term from the beginning.
i.e., \(n^{\text {th }}\) term from the end
\(
\begin{aligned}
& =(m-n+1)^{\text {th }} \text { term from the beginning } \\
& =a+(m-n+1-1) d=a+(m-n) d
\end{aligned}
\)
\(n^{\text {th }} \text { term from the end }\):Â We can also compute \(n^{\text {th }}\) term from the end by taking the last term as first term and common difference as ‘ \(-d\)‘.
Thus, if \(l\) is the last term. Then,
\(
n^{\text {th }} \text { term from the end }=l+(n-1)(-d)=l-(n-1) d
\)
Example 2: Prove that in any A.P., the sum of terms equidistant from the beginning and the end is constant and equal to twice the middle term (if one exists).
Solution: This property is one of the most useful “shortcuts” in Arithmetic Progressions. It states that in a finite A.P., the sum of any two terms that are the same distance from the start and the end will always be the same.
The Core Property:
If you have an A.P. with \(n\) terms: \(a_1, a_2, a_3, \ldots, a_{n-1}, a_n\), then:
\(
a_1+a_n=a_2+a_{n-1}=a_3+a_{n-2}=\cdots=\text { Constant }
\)
Example with an Even Number of Terms
Let’s look at the A.P.: 2, 5, 8, 11, 14, 17 (where \(d=3\) )
1st and 6th (ends): \(2+17=19\) (constant), Also sum of 1st (2) and 3rd (8) term is \(2 \times \) middle term (5) (\(2 + 8 = 10 = 2 \times 5\)).
2nd and 5th: \(5+14=19\) (constant), Also sum of 2nd (5) and 4th (11) term is \(2 \times \) middle term (8) (\(5 + 11 = 16 = 2 \times 8\)).
3rd and 4th: \(8+11=19\) (constant), Also sum of 3rd (8) and 5th (14) term is \(2 \times \) middle term (11) (\(8 + 14 = 22 = 2 \times 11\)).
Since the sum is always 19, you can see the symmetry.
Why does this work? In an A.P., as you move one step “in” from the left, you add \(d\). As you move one step “in” from the right, you subtract \(d\). These changes cancel each other out, keeping the sum constant.
Example 3: Show that the sequence \(9,12,15,18, \ldots\) is an A.P. Find its \(\mathbf{1 6}^{\text {th }}\) term and the general term.
Solution: Since \((12-9)=(15-12)=(18-15)=3\), therefore the given sequence is an A.P. with common difference 3. First term is 9. Therefore, the \(16^{\text {th }}\) term is
\(
\begin{aligned}
a_{16} & =a+(16-1) d \quad\left[\because a_n=a+(n-\mathbf{1}) d\right] \\
& =a+15 d \\
\Rightarrow \quad a_{16} & =9+15 \times 3=54
\end{aligned}
\)
The general term ( \(n^{\text {th }}\) term) is given by
\(
\begin{aligned}
a_n & =a+(n-1) d \\
& =9+(n-1) \times 3=3 n+6
\end{aligned}
\)
Example 4: How many terms are there in the A.P. 3, 7, 11, …, 407?
Solution: We know that last term \(a_n=a+(n-1) d\)
Where \(d=\) common difference \(=4\)
and \(\mathrm{a}=\) first term \(=3\)
\(
\begin{aligned}
407 & =3+(n-1) 4 \\
\Rightarrow \quad n & =102
\end{aligned}
\)
Hence there are 102 terms in A.P.
Example 5: If \(a, b, c, d, e\) are in A.P., then find the the value of \(a-4 b+6 c-4 d+e\).
Solution: Key Concept: Equidistant Terms
In any A.P., the sum of terms equidistant from the beginning and the end is constant and equal to twice the middle term (if one exists).
Breakdown of the Solution:
Step 1: Regrouping for Symmetry: The expression \(E=a-4 b+6 c-4 d+e\) is rearranged to group terms that are “balanced” around the center term \(c\) :
\(
E=(a+e)-4(b+d)+6 c
\)
Step 2: Using the Mean Property:
For \(b, c, d\) : Since \(c\) is exactly in the middle of \(b\) and \(d\), their average is \(c\).
\(
\frac{b+d}{2}=c \Longrightarrow b+d=2 c
\)
For \(a, c, e\) : Similarly, \(c\) is the middle term between \(a\) and \(e\).
\(
\frac{a+e}{2}=c \Longrightarrow a+e=2 c
\)
Step 3: Substitution and Simplification: Replace \((a+e)\) and \((b+d)\) with \(2 c\) in the regrouped equation:
\(
\begin{gathered}
E=(2 c)-4(2 c)+6 c \\
E=2 c-8 c+6 c \\
E=0
\end{gathered}
\)
Example 5: In a certain A.P., 5 times the \(5^{\text {th }}\) term is equal to 8 times the \(8^{\text {th }}\) term, then prove that its \(13^{\text {th }}\) term is \(\mathbf{0}\).
Solution: The general term of an A.P. is \(T_n=a+(n-1) d\).
Given, \(5 T_5=8 T_8\)
\(T_5=a+4 d\)
\(T_8=a+7 d\).
\(
5(a+4 d)=8(a+7 d)
\)
\(
a=-12 d
\)
\(
T_{13}=a+(13-1) d=a+12 d=(-12 d)+12 d=0
\)
Example 6: Find the term of the series 25 , \(22 \frac{3}{4}, 20 \frac{1}{2}, 18 \frac{1}{4}, \ldots\) which is numerically the smallest.
Solution: This problem is interesting because it asks for the numerically smallest term. This means we are looking for the value closest to zero, regardless of whether it is positive or negative (the absolute value \(\left|T_n\right|\) should be minimum).
The given series is an A.P.
Step-by-Step Breakdown:
Step 1: Identify the A.P. Parameters
First term (a): 25
Common difference \((d): 22 \frac{3}{4}-25=-2 \frac{1}{4}=-\frac{9}{4}\)
Since \(d\) is negative, the terms are decreasing and will eventually cross zero and become negative.
Step 2: Find the General Term ( \(T_n\) ) Using the formula \(T_n=a+(n-1) d\) :
\(
\begin{gathered}
T_n=25+(n-1)\left(-\frac{9}{4}\right) \\
T_n=25-\frac{9}{4} n+\frac{9}{4} \\
T_n=\frac{100+9}{4}-\frac{9}{4} n=\frac{109-9 n}{4}
\end{gathered}
\)
Step 3: Determine where the sequence crosses Zero To find the transition point, set \(T_n<0\) :
\(
\begin{gathered}
\frac{109-9 n}{4}<0 \Longrightarrow 109<9 n \Longrightarrow n>\frac{109}{9} \\
n>12 \frac{1}{9}
\end{gathered}
\)
This tells us:
The \(12^{\text {th }}\) term is the last positive term.
The \(13^{\text {th }}\) term is the first negative term.
Step 4: Compare the Candidates for “Smallest”
The numerically smallest term must be either the last positive one or the first negative one. Let’s calculate both:
\(
\begin{array}{llll}
\text { Term } & \text { Calculation } & \text { Value } & \text { Absolute Value } \\
T_{12} & \frac{109-9(12)}{4}=\frac{109-108}{4} & \frac{1}{4}(0.25) & 0.25 \\
T_{13} & \frac{109-9(13)}{4}=\frac{109-117}{4} & -2 & 2
\end{array}
\)
Conclusion: Since 0.25 is closer to zero than -2 is, the \(12^{\text {th }}\) term ( \(1 / 4\) ) is the numerically smallest term in the series.
Example 7: Given two A.P.’s \(2,5,8,11, \ldots \ldots, T_{60}\) and \(3,5,7,9, \ldots \ldots . T_{50}\). Then find the number of terms which are identical.
Solution: To find the number of identical terms between the two Arithmetic Progressions (A.P.s), we follow these steps:
Step 1: Analyze the first A.P.
The first A.P. is \(2,5,8,11, \ldots\)
First term ( \(a_1\) ): 2
Common difference \(\left(d_1\right): 5-2=3\)
Number of terms \(\left(n_1\right)\) : 60
Last term ( \(T_{60}\) ):
Step 2: Analyze the second A.P.
The second A.P. is \(3,5,7,9, \ldots\)
First term ( \(a_2\) ): 3
Common difference \(\left(d_2\right): 5-3=2\)
Number of terms ( \(n_2\) ): 50
Last term ( \(T_{50}^{\prime}\) ):
\(
T_{50}=a_2+(50-1) d_2=3+49 \times 2=3+98=101
\)
Step 3: Find the identical terms
The common terms will themselves form an A.P.
First common term ( \(a_c\) ): By inspection, the first identical term is 5.
Common difference of identical terms ( \(d_c\) ): This is the Least Common Multiple (L.C.M.) of the common differences of the two A.P.’s.
\(
d_c=\text { L.C.M. }\left(d_1, d_2\right)=\text { L.C.M. }(3,2)=6
\)
Also common terms are are \(5,11, \ldots\)
Step 4: Determine the number of identical terms
The common terms must not exceed the last term of the shorter sequence. The last term of the first sequence is 179 and the last term of the second is 101. Thus, the maximum possible value for a common term is 101.
Let the number of identical terms be \(n\). The \(n\)-th common term is given by:
\(
T_n=a_c+(n-1) d_c
\)
We set the condition:
\(
\begin{gathered}
5+(n-1) 6 \leq 101 \\
(n-1) 6 \leq 96
\end{gathered}
\)
\(
\begin{gathered}
n-1 \leq 16 \\
n \leq 17
\end{gathered}
\)
The largest integer value for \(n\) is 17.
Conclusion: The number of terms which are identical is 17. Hence 101 is the actual last common term.
Example 8: Consider two A.P.’s:
\(S_1: 2,7,12,17, \ldots 500\) terms
and \(\quad S_2: 1,8,15,22, \ldots 300\) terms
Find the number of common terms. Also find the last common term.
Solution: \(S_1: 2,7,12,17, \ldots 500\) terms
\(
\begin{array}{ll}
\Rightarrow & T_{500}=2+(500-1) 5=2497 \\
& S_2: 1,8,15,22, \ldots 300 \text { terms } \\
\Rightarrow & T_{300}=1+(300-1) 7=2094
\end{array}
\)
Common differences of \(S_1\) and \(S_2\) are 5 and 7 respectively.
Hence common difference of common term series is 35 (this is the L.C.M. of common differences of \(S_1\) and \(S_2\) that is \(5\) and \(7\).)
A.P. of common terms is \(22,57,92, \ldots\)
The common terms must be less than or equal to the smaller of the two last terms. Maximum \(\operatorname{limit}\left(L_{\max }\right)=\min (2497,2094)=2094\).
Let last term is \(2094 \Rightarrow 22+(n-1) 35=2094 \Rightarrow n=60.2\)
But \(\boldsymbol{n}\) is natural number \(\Rightarrow n=60\) (number of common terms)
Then actual last common term \(=22+(60-1) 35=2087\)
Some Important Facts about A.P.
Property 1: If a fixed number is added or subtracted to each term of a given A.P., then the resulting series is also an A.P., and its common difference remains the same.
Illustration 1: Adding a positive number
Consider the A.P.: \(5,10,15,20, \ldots\) (Common difference \(d=5\) ) Let’s add 3 to each term:
\(5+3=8\)
\(10+3=13\)
\(15+3=18\)
\(20+3=23\)
New Sequence: 8, 13, 18, 23, . . Verification: \(13-8=5\) and \(18-13=5\). The common difference is still 5.
Illustration 2: Subtracting a number
Consider the A.P.: 20, 18, 16, 14, . . (Common difference \(d=-2\) ) Let’s subtract 10 from each term:
\(20-10=10\)
\(18-10=8\)
\(16-10=6\)
\(14-10=4\)
New Sequence: \(10,8,6,4, \ldots\) Verification: \(8-10=-2\) and \(6-8=-2\). The common difference is still -2.
Property 2: If each term of an A.P. is multiplied by a fixed constant or divided by a fixed non-zero constant, then the resulting series in also an A.P.
Illustration 3: Multiplication
Consider the A.P.: 3, \(6,9,12, \ldots\) (Common difference \(d=3\) ) Let’s multiply by 2:
\(3 \times 2=6\)
\(6 \times 2=12\)
\(9 \times 2=18\)
\(12 \times 2=24\)
New Sequence: \(6,12,18,24, \ldots\) Verification: \(12-6=6\) and \(18-12=6\). The new common difference is 6 (which is \(3 \times 2\) ).
Illustration 4: Division
Consider the A.P.: 10,20,30,40, \(\ldots\) (Common difference \(d=10\) ) Let’s divide by 5 :
\(10 \div 5=2\)
\(20 \div 5=4\)
\(30 \div 5=6\)
\(40 \div 5=8\)
New Sequence: \(2,4,6,8, \ldots\) Verification: \(4-2=2\) and \(6-4=2\). The new common difference is 2 (which is \(10 \div 5\)).
Property 3: If \(x_1+x_2+x_3+\cdots\) and \(y_1+y_2+y_3+\cdots\) are two A.P.’s, then \(x_1 \pm y_1, x_2 \pm y_2, x_3 \pm y_3, \ldots\) are also A.P.’s.
Illustration 5: Term-by-Term Addition
Consider the following two sequences:
\(X: 2,5,8,11, \ldots\left(d_x=3\right)\)
\(Y: 10,20,30,40, \ldots\left(d_y=10\right)\)
Adding them ( \(x_n+y_n\) ):
\(2+10=12\)
\(5+20=25\)
\(8+30=38\)
\(11+40=51\)
Resulting Sequence: 12, 25, 38, 51, . . . Verification:
\(25-12=13\)
\(38-25=13\)
Note that \(d_x+d_y=3+10=13\).
Illustration 6: Term-by-Term Subtraction
Using the same sequences as above, let’s calculate \(y_n-x_n\) :
\(10-2=8\)
\(20-5=15\)
\(30-8=22\)
\(40-11=29\)
Resulting Sequence: 8, 15, 22, 29, . . . Verification:
\(15-8=7\)
\(22-15=7\)
Note that \(d_y-d_x=10-3=7\).
Property 4: Three terms in an A.P. should preferably be taken as \(a-d, a, a+d\) and four terms as \(a-3 d, a-d, a+d, a +3 d\).
Illustration 7: This specific selection of terms simplifies calculations, particularly when the sum of the terms is given, because the common difference \({d}\) cancels out when the terms are added together.
For three terms in A. P.: the sum is \((a-d)+a+(a+d)=3 a\).
For four terms in A. P.: the sum is \((a-3 d)+(a-d)+(a+d)+(a+3 d)=4 a\).
This immediately allows one to find the value of \(a\), the middle term (or the value related to the middle terms), making the rest of the problem easier to solve.
Property 5: In A.P., \(a_n=\frac{a_{n-k}+a_{n+k}}{2}\), for \(k \leq n\).
Finding a Middle Term:
Illustration 8: In an A.P., the \(7^{\text {th }}\) term is 14 and the \(13^{\text {th }}\) term is 30. What is the \(10^{\text {th }}\) term?
Solve the system: \(a_7=a+6 d=14\) and \(a_{13}=a+12 d=30\).
Find \(d=\frac{16}{6}=\frac{8}{3}\).
Find \(a=14-6\left(\frac{8}{3}\right)=-2\).
Calculate \(a_{10}=-2+9\left(\frac{8}{3}\right)=22\).
The Shortcut (Using your property): Notice that the index 10 is exactly halfway between 7 and 13 (where \(k=3\) ).
\(
\begin{gathered}
a_{10}=\frac{a_{10-3}+a_{10+3}}{2}=\frac{a_7+a_{13}}{2} \\
a_{10}=\frac{14+30}{2}=22
\end{gathered}
\)
Verifying a Sequence:
Illustration 9: Is the sequence \(15, x, 25,30,35\) an A.P. if \(x=20\) ?
Application: Pick any term and check if it’s the average of its neighbors. Let’s look at the term \(25\left(a_3\right)\). Its neighbors are \(x\left(a_2\right)\) and \(30\left(a_4\right)\).
\(
\begin{gathered}
a_3=\frac{a_2+a_4}{2} \\
25=\frac{20+30}{2}=\frac{50}{2}=25
\end{gathered}
\)
The property holds for \(k=1\). You can also check \(k=2\) using the first and fifth terms:
\(
\frac{a_1+a_5}{2}=\frac{15+35}{2}=\frac{50}{2}=25
\)
Since the middle term is the average of equidistant terms, it confirms the sequence is an A.P.
Property 6:Â It \(a, a_2, a_3, \ldots a_n\) are in A.P. Then \(a_1+a_n=a_2+a_{n-1}=a_3 +a_{n-2}=\ldots=a_r+a_{n-r+1}\).
Illustration 10: Simple Integer Sequence:
Consider the first 6 terms of an A.P. with \(a=2\) and \(d=3\) : Sequence: \(2,5,8,11,14,17\)
1st+6th: \(2+17=19\)
2nd + 5th: \(5 + 14 = 19\)
3rd + 4th: \(8+11=19\)
Illustration 11: Finding a Sum with Limited Info:
In an A.P. of 20 terms, if the \(3^{\text {rd }}\) term is 10 and the \(18^{\text {th }}\) term is 40 , what is the sum of the \(1^{\text {st }}\) and \(20^{\text {th }}\) terms?
Solution: Using the property, we know that \(a_3\) and \(a_{18}\) are equidistant from the ends because their indices add up to \(n+1(3+18=21\) and \(1+20=21)\). Therefore:
\(
\begin{gathered}
a_1+a_{20}=a_3+a_{18} \\
a_1+a_{20}=10+40=50
\end{gathered}
\)
Illustration 12: Solving Complex Summations: If \(a_1, a_2, \ldots, a_{24}\) are in A.P. and \(a_1+a_5+a_{10}+a_{15}+a_{20}+a_{24}=300\), find the sum of the first 24 terms ( \(S_{24}\) ).
Solution: Pair the terms equidistant from the ends:
\(\left(a_1+a_{24}\right)\)
\(\left(a_5+a_{20}\right), \) Note: \(5+20=25\), which is \(n+1\)
\(\left(a_{10}+a_{15}\right), \) Note: \(10+15=25\)
All these pairs are equal to \(\left(a_1+a_{24}\right)\). Let \(S=a_1+a_{24}\).
The equation becomes: \(S+S+S=300 \Longrightarrow 3 S=300 \Longrightarrow S=100\).
The formula for the sum of \(n\) terms is \(S_n=\frac{n}{2}\left(a_1+a_n\right)\).
\(S_{24}=\frac{24}{2}(100)=12 \times 100=1200\).
Example 9: The sum of three numbers in A.P. is -3 and their product is \(\mathbf{8}\). Find the numbers.
Solution: Let the numbers be \((a-d), a,(a+d)\). Therefore,
\(
\begin{aligned}
& (a-d)+a+(a+d)=-3 \\
\Rightarrow & 3 a=-3 \\
\Rightarrow & a=-1
\end{aligned}
\)
\(
\begin{aligned}
& (a-d)(a)(a+d)=8 \\
\Rightarrow & a\left(a^2-d^2\right)=8 \\
\Rightarrow & (-1)\left(1-d^2\right)=8 \quad[\because a=-1] \\
\Rightarrow & d^2=9 \\
\Rightarrow & d= \pm 3
\end{aligned}
\)
If \(d=3\), the numbers are \(-4,-1,2\). If \(d=-3\), the numbers are 2 , \(-1,-4\). So, the numbers are \(-4,-1,2\) or \(2,-1,-4\).
Example 10: Divide 32 into four parts which are in A.P. such that the ratio of the product of extremes to the product of means is 7:15.
Solution: Let the four parts be \((a-3 d),(a-d),(a+d)\) and \((a+3 d)\). Then,
\(
\begin{aligned}
& (a-3 d)+(a-d)+(a+d)+(a+3 d)=32 \\
\Rightarrow & 4 a=32 \\
\Rightarrow & a=8
\end{aligned}
\)
Also,
\(
\begin{aligned}
& \frac{(a-3 d)(a+3 d)}{(a-d)(a+d)}=\frac{7}{15} \\
\Rightarrow & \frac{a^2-9 d^2}{a^2-d^2}=\frac{7}{15} \\
\Rightarrow & \frac{64-9 d^2}{64-d^2}=\frac{7}{15} \\
\Rightarrow & 128 d^2=512 \\
\Rightarrow & d^2=4
\end{aligned}
\)
\(
\Rightarrow d= \pm 2
\)
Thus, the four parts are \(2,6,10,14\).
Example 11: The digits of a positive integer, having three digits, are in A.P. and their sum is 15. The number obtained by reversing the digits is 594 less than the original number. Find the number.
Solution: Let the digits at ones, tens and hundreds place be ( \(a-d\) ), \(a\) and ( \(a+d\) ), respectively. Then the number is
\(
(a+d) \times 100+a \times 10+(a-d)=111 a+99 d
\)
The number obtained by reversing the digits is.
\(
(a-d) \times 100+a \times 10+(a+d)=111 a-99 d
\)
It is given that
\(
(a-d)+a+(a+d)=15
\)
and
\(
\begin{aligned}
& 111 a-99 d=111 a+99 d-594 \\
\therefore & 3 a=15 \text { and } 198 d=594 \\
\Rightarrow & a=5 \text { and } d=3
\end{aligned}
\)
So, the number is \(111 \times 5+99 \times 3=852\).
Sum of \(n\) terms of an A.P.
The sum \(S_n\) of \(n\) terms of an A.P. with the first term ‘ \(a\) ‘ and the common difference ‘ \(d\) ‘ is
\(
S_n=\frac{n}{2}[2 a+(n-1) d] \text { or } S_n=\frac{n}{2}[a+l]
\)
where \(l=\) last term \(=a+(n-1) d\).
Proof: The sum can be written in two ways:
\(
\begin{aligned}
& S_n=a_1+a_2+a_3+\cdots+a_{n-2}+a_{n-1}+a_n \\
& S_n=a_n+a_{n-1}+a_{n-2}+\cdots+a_3+a_2+a_1
\end{aligned}
\)
Adding corresponding terms in these two equations, we get \(2 S_n\) as the sum of \(n\) pairs, each equal to \(\left(a_1+a_n\right)\) :
\(
\begin{gathered}
2 S_n=\left(a_1+a_n\right)+\left(a_2+a_{n-1}\right)+\cdots+\left(a_{n-1}+a_2\right)+\left(a_n+a_1\right) \\
2 S_n=n\left(a_1+a_n\right)
\end{gathered}
\)
This yields the first form of the sum formula:
\(
S_n=\frac{n}{2}\left(a_1+a_n\right)
\)
Substituting \(a_n=a_1+(n-1) d\) (using ‘ \(a\) ‘ for the first term as originally defined):
\(
S_n=\frac{n}{2}\{a+(a+(n-1) d)\}
\)
Which simplifies to the second standard form:
\(
S_n=\frac{n}{2}[2 a+(n-1) d]
\)
Example 12:Â If the sum of the series \(2,5,8,11, \ldots\) is 60100 , then find the value of \(n\).
Solution: Here first term is \(a=2\) and common difference \(d=3\) Hence sum of \(n\) terms of A.P. is, \((n / 2)(4+3(n-1))=60100\)
\(
\begin{aligned}
& \Rightarrow \quad 3 n^2+n-120200=0 \\
& \Rightarrow \quad(n-200)(3 n+601)=0 \\
& \Rightarrow \quad n=200
\end{aligned}
\)
Example 13: In an A.P. if \(S_1=T_1+T_2+T_3+\cdots+T_n \left(n\right.\) odd), \(S_2=T_2+T_4+T_6+\cdots+T_{n-1}\), then find the value of \(S_1 / S_2\) in terms of \(n\).
Solution: This is a classic problem that tests your understanding of how an A.P. can be split into smaller sub-sequences.
Step 1: Analyzing \(S_1\) (The Full Series)
\(S_1\) is the sum of all \(n\) terms. Since \(n\) is odd, we can use the formula:
\(
S_1=\frac{n}{2}\left(T_1+T_n\right)
\)
Step 2: Analyzing \(S_2\) (The Even-Positioned Terms)
\(S_2\) consists of terms at the 2 nd, 4th, 6th… \((\mathrm{n}-1)\)th positions.
Number of terms: Since \(n\) is odd (let \(n=2 m+1\) ), the number of even positions is exactly \(\frac{n-1}{2}\).
First term of \(S_2: T_2\)
Last term of \(S_2: T_{n-1}\)
Using the sum formula \(S=\frac{\text { number of terms }}{2}\) (first + last):
\(
S_2=\frac{\frac{n-1}{2}}{2}\left(T_2+T_{n-1}\right)=\frac{n-1}{4}\left(T_2+T_{n-1}\right)
\)
Step 3: Applying the “Constant Pair Sum” Property
From the property we discussed earlier ( \(a_1+a_n=a_2+a_{n-1}\) ), we know:
\(
T_1+T_n=T_2+T_{n-1}
\)
Let this constant sum be \(K\). Now we can rewrite the sums:
\(S_1=\frac{n}{2} \cdot K\)
\(S_2=\frac{n-1}{4} \cdot K\)
Step 4: Finding the Ratio \(S_1 / S_2\)
\(
\frac{S_1}{S_2}=\frac{\frac{n}{2} \cdot K}{\frac{n-1}{4} \cdot K}
\)
The \(K\) values cancel out:
\(
\frac{S_1}{S_2}=\frac{n}{2} \div \frac{n-1}{4}
\)
\(
\begin{gathered}
\frac{S_1}{S_2}=\frac{n}{2} \times \frac{4}{n-1} \\
\frac{S_1}{S_2}=\frac{2 n}{n-1}
\end{gathered}
\)
Alternative Method: The “Average” Approach
There is another quick way to verify your result using the concept of the Middle Term.
For any A.P. with \(n\) terms, the sum \(S_1=n \times\) (Average).
The average of an A.P. is the middle term, \(T_{\text {mid }}\)
In \(S_2\), the terms are \(T_2, T_4, \ldots, T_{n-1}\). These terms are also an A.P. symmetrically distributed around the same middle term \(T_{\text {mid }}\).
Therefore, \(S_2=\) (number of terms in \(\left.S_2\right) \times T_{\text {mid }}\).
As you calculated, the number of terms in \(S_2\) is \(\frac{n-1}{2}\).
\(
\frac{S_1}{S_2}=\frac{n \times T_{m i d}}{\frac{n-1}{2} \times T_{m i d}}=\frac{n}{\frac{n-1}{2}}=\frac{2 n}{n-1}
\)
Example 14: In an A.P. of 99 terms, the sum of all the odd numbered terms is \(\mathbf{2 5 5 0}\). Then find the sum of all the 99 terms of the A.P.
Solution: The Middle Term Approach (Fastest)
For any A.P. with an odd number of terms, the sum is simply the middle term multiplied by the total number of terms.
Find the middle term ( \(a_{50}\) ): The odd-indexed terms are \(a_1, a_3, a_5, \ldots, a_{99}\). There are \(\frac{99+1}{2}=50\) such terms. The middle of these 50 terms is the average of the first and last index: \(\frac{1+99}{2}=50\). So, the sum of these 50 terms is:
\(
\begin{aligned}
& 50 \times a_{50}=2550 \\
& a_{50}=\frac{2550}{50}=51
\end{aligned}
\)
Find the total sum ( \(S_{99}\) ): The middle term of the entire 99-term sequence is also \(a_{50}\)
\(
\begin{gathered}
S_{99}=99 \times a_{50} \\
S_{99}=99 \times 51=5049
\end{gathered}
\)
Alternate Method: The Sum of Extremes Approach
Sum of odd-numbered terms: The first term is \(a_1\) and the last (99th) is \(a_{99}\).
\(
\begin{gathered}
S_{\text {odd }}=\frac{50}{2}\left(a_1+a_{99}\right)=2550 \\
25\left(a_1+a_{99}\right)=2550 \\
a_1+a_{99}=\frac{2550}{25}=102
\end{gathered}
\)
Sum of all 99 terms:
\(
S_{99}=\frac{99}{2}\left(a_1+a_{99}\right)
\)
Substitute the value \(a_1+a_{99}=102\) :
\(
S_{99}=\frac{99}{2}(102)=99 \times 51=5049
\)
Example 15: Find the degree of the expression \((1+x) \left(1+x^6\right)\left(1+x^{11}\right) \ldots\left(1+x^{101}\right)\).
Solution: The degree of the expression is \(1+6+11+\cdots+101\) which is an A.P.
Now \(101=1+5(\mathrm{n}-1) \Rightarrow n=21\)
\(
\begin{aligned}
\Rightarrow & 1+6+11+\ldots+101 \\
= & \frac{21}{2}[1+101]=21 \times 51=1071
\end{aligned}
\)
Example 16: Find the sum of all three-digit natural numbers, which are divisible by 7.
Solution: The smallest and the largest numbers of three digits, which are divisible by 7 are 105 and 994 , respectively. So, the sequence of three digit numbers which are divisible by 7 is \(105,112,119, \ldots, 994\). Clearly, it is an A.P. with first term \(a =105\) and common difference \(d=7\). Let there be \(n\) terms in this sequence. Then,
\(
\begin{aligned}
& a_n=994 \\
\Rightarrow & a+(n-1) d=994 \\
\Rightarrow & 105+(n-1) \times 7=994 \\
\Rightarrow & n=128
\end{aligned}
\)
Now, required sum is
\(
\begin{aligned}
& \frac{n}{2}[2 a+(n-1) d] \\
& =\frac{128}{2}[2 \times 105+(128-1) \times 7] \\
& =70336
\end{aligned}
\)
Example 17: Let \(T_r\) be the \(r^{\text {th }}\) term of an AP, for \(r=1,2, \ldots\) If for some positive integers \(m, n\) we have \(T_m=1 / n\) and \(T_n=1 / m\), then \(T_{m n}\) equals [IIT 1998]
(a) \(\frac{1}{m n}\)
(b) \(\frac{1}{m}+\frac{1}{n}\)
(c) 1
(d) 0
Solution: Step 1: Set up the Equations
Let the first term be \(a\) and the common difference be \(d\). Using the formula \(T_r=a+(r-1) d\)
\(T_m=a+(m-1) d=\frac{1}{n} \dots(1)\)
\(T_n=a+(n-1) d=\frac{1}{m} \dots(2)\)
Step 2: Solve for \(d\)
Subtract equation (2) from equation (1):
\(
\begin{gathered}
(a+(m-1) d)-(a+(n-1) d)=\frac{1}{n}-\frac{1}{m} \\
(m-1-n+1) d=\frac{m-n}{m n} \\
(m-n) d=\frac{m-n}{m n}
\end{gathered}
\)
Since \(m \neq n\), we can divide by ( \(m-n\) ):
\(
d=\frac{1}{m n}
\)
Step 3: Solve for \(a\)
Substitute the value of \(d\) back into equation (1):
\(
\begin{gathered}
a+(m-1)\left(\frac{1}{m n}\right)=\frac{1}{n} \\
a+\frac{m}{m n}-\frac{1}{m n}=\frac{1}{n} \\
a+\frac{1}{n}-\frac{1}{m n}=\frac{1}{n} \\
a=\frac{1}{m n}
\end{gathered}
\)
Step 4: Find \(T_{m n}\)
Now, use the values of \(a\) and \(d\) to find the \(m n^{\text {th }}\) term:
\(
\begin{gathered}
T_{m n}=a+(m n-1) d \\
T_{m n}=\frac{1}{m n}+(m n-1)\left(\frac{1}{m n}\right) \\
T_{m n}=\frac{1}{m n}+\frac{m n}{m n}-\frac{1}{m n} \\
T_{m n}=1
\end{gathered}
\)
Example 19: Let \(T_r\) be the \(r^{\text {th }}\) term of an A.P. whose first term is \(a\) and common difference is \(d\). If for some positive integers \(m, n\), \(m \neq n, T_m=\frac{1}{n}\) and \(T_n=\frac{1}{m}\), then \(a-d\) equals [AIEEE 2004]
(a) \(\frac{1}{m}+\frac{1}{n}\)
(b) 1
(c) \(\frac{1}{m n}\)
(d) 0
Solution: Set up the Equations
Let the first term be \(a\) and the common difference be \(d\). Using the formula \(T_r=a+(r-1) d\)
\(T_m=a+(m-1) d=\frac{1}{n} \dots(1)\)
\(T_n=a+(n-1) d=\frac{1}{m} \dots(2)\)
We established that when \(T_m=\frac{1}{n}\) and \(T_n=\frac{1}{m}\) :
Common Difference ( \(d\) ): Subtracting the two equations \((a+(m-1) d)-(a+(n-\) 1) \(d)=\frac{1}{n}-\frac{1}{m}\) led to:
\(
\begin{aligned}
&\begin{gathered}
(m-1-n+1) d=\frac{m-n}{m n} \\
(m-n) d=\frac{m-n}{m n}
\end{gathered}\\
&\text { Since } m \neq n \text {, we can divide by }(m-n) \text { : }\\
&d=\frac{1}{m n}
\end{aligned}
\)
First Term ( \(a\) ): Substituting \(d\) back into \(a+(m-1) d=\frac{1}{n}\) :
\(
\begin{gathered}
a+(m-1) \frac{1}{m n}=\frac{1}{n} \\
a=\frac{1}{n}-\frac{m-1}{m n}=\frac{m-(m-1)}{m n}=\frac{1}{m n}
\end{gathered}
\)
Calculating \(a-d\)
Now, we simply subtract the two values:
\(
a-d=\frac{1}{m n}-\frac{1}{m n}=0
\)
Some Useful Results on the Sum of an A.P.
Result-1: A sequence is an A.P. iff the sum of its \(n\) terms is of the form \(A n^2+B n\), where \(A, B\) are constants. In such a case, the common difference is equal to \(2 A\).
Example 20: If \(S_n=n P+\frac{n(n-1)}{2} Q\), where \(S_n\) denotes the sum of the first \(n\) terms of an \(A . P\)., then the common difference is [JEE (WB) 1994, CEE (Delhi) 2001]
(a) \(P+Q\)
(b) \(2 P+3 Q\)
(c) \(2 Q\)
(d) \(Q\)
Solution: (d) Let \(a_n\) be the \(n^{\text {th }}\) term of the A.P. Then,
\(
\begin{aligned}
& a_n=S_n-S_{n-1} \\
\Rightarrow \quad & a_n=\left\{n P+\frac{n(n-1)}{2} Q\right\}-\left\{(n-1) P+\frac{(n-1)(n-2)}{2} Q\right\} \\
\Rightarrow \quad & a_n=P+(n-1) Q \\
\therefore \quad \text { Common difference } & =a_n-a_{n-1} \\
& =[P+(n-1) Q]-[P+(n-2) Q]=Q
\end{aligned}
\)
ALITER: Common difference \(=2\left(\right.\) Coeff. of \(\left.n^2\right)=Q\)
Example 21: If the fourth power of the common difference of an A.P. with integer entries is added to the product of any four consecutive terms of it, then the resulting sum is [IIT 2000]
(a) an even integer
(b) an odd integer
(c) the square of an integer
(d) the cube of an integer
Solution: (c) Let \(a-3 d, a-d, a+d, a+3 d\) be four consecutive terms of an A.P. with integer terms. Clearly, the common difference is \(2 d\).
Since the terms are integers, therefore \(a\) and \(d\) are also integers. Also,
\(
\begin{aligned}
& (a-3 d)(a-d)(a+d)(a+3 d)+(2 d)^4 \\
& =\left(a^2-9 d^2\right)\left(a^2-d^2\right)+16 d^4 \\
& =a^4-10 a^2 d^2+9 d^4+16 d^4 \\
& =a^4-10 a^2 d^2+25 d^4 \\
& =\left(a^2-5 d^2\right)^2, \text { which is square of an integer as } a \text { and } d \text { are integers.}
\end{aligned}
\)
Result-2: If the ratio of \(n^{\text {th }}\) terms of two A.P.s is given, then the ratio of the sums of their \(n\) terms may be found by replacing \(n\) by \(\frac{n+1}{2}\) in the ratio of the \(n^{\text {th }}\) terms.
Example 22: If the ratio of \(n^{\text {th }}\) terms of two A.P.s is \((2 n+8):(5 n-3)\), then the ratio of the sums of their \(n\) terms is
(a) \((2 n+18):(5 n+1)\)
(b) \((5 n-1):(2 n+18)\)
(c) \((2 n+18):(5 n-1)\)
(d) none of these
Solution: (c) Required ratio is obtained by replacing \(n\) by \(\frac{n+1}{2}\)
\(
\begin{aligned}
\therefore \quad \text { Required ratio } & =\left\{\frac{2(n+1)}{2}+8\right\}:\left\{\frac{5(n+1)}{2}-3\right\} \\
& =(2 n+18):(5 n-1)
\end{aligned}
\)
Example 23: Let \(a_1, a_2, a_3, \ldots\) be terms of an A.P. If \(\frac{a_1+a_2+\ldots+a_p}{a_1+a_2+\ldots+a_q}=\frac{p^2}{q^2}, p \neq q\), then \(\frac{a_6}{a_{21}}\) equals [AIEEE 2006]
(a) \(\frac{41}{11}\)
(b) \(\frac{7}{2}\)
(c) \(\frac{2}{7}\)
(d) \(\frac{11}{41}\)
Solution: (d) We have,
\(
\begin{aligned}
& \frac{a_1+a_2+\ldots+a_p}{a_1+a_2+\ldots+a_q}=\frac{p^2}{q^2} \\
\Rightarrow & \frac{\frac{p}{2}\left\{2 a_1+(p-1) d\right\}}{\frac{q}{2}\left\{2 a_1+(q-1) d\right\}}=\frac{p^2}{q^2} \Rightarrow \frac{a_1+\left(\frac{p-1}{2}\right) d}{a_1+\left(\frac{q-1}{2}\right) d}=\frac{p}{q} \dots(i)
\end{aligned}
\)
We have to find the value of \(\frac{a_6}{a_{21}}\) i.e. \(\frac{a_1+5 d}{a_1+20}\).
So, replacing \(\frac{p-1}{2}\) by 5 and \(\frac{q-1}{2}=20\) i.e. putting \(p=11\) and \(q=41\) in (i), we get \(\frac{a_6}{a_{12}}=\frac{11}{41}\).
Arithmetic Means
If between \(a\) and \(b\), two given quantities, we have to insert \(n\) quantities \(A_1, A_2, \ldots, A_n\) such that \(a, A_1, A_2, \ldots, A_n, b\) forms an A.P., then we say that \(A_1, A_2, \ldots, A_n\) are known as \(n\) arithmetic means between \(a\) and \(b\).
For example, 15, 11, 7, 3, -1, -5 are in A.P. It follows that \(11,7,3,-1\) are four arithmetic means between 15 and -5 with common difference \(d=-4\) (\(d=\frac{b-a}{n+1}\) Where, \(n=4\) (the number of means inserted)).
If \(a, A, b\) are in A.P., we say that \(A\) is the arithmetic mean between \(a\) and \(b\). When we only insert one mean between \(a\) and \(b\), that mean \(A\) is simply the average of the two numbers.
If \(a, A, b\) are in A.P., then:
\(
\begin{gathered}
A-a=b-A \\
2 A=a+b
\end{gathered}
\)
\(
\begin{aligned}
&A=\frac{a+b}{2}\\
&\text { For example, the single arithmetic mean between } 15 \text { and -5 is: }\\
&\frac{15+(-5)}{2}=\frac{10}{2}=5
\end{aligned}
\)
Insertion of \(n\) Arithmetic Means Between \(a\) and \(b\)
Let \(A_1, A_2, \ldots, A_n\) be \(n\) arithmetic means between two quantities \(a\) and \(b\). Then, \(a, A_1, A_2, \ldots, A_n, b\) is an A.P. Let \(d\) be the common difference of this A.P. Clearly, it contains \(n+2\) terms.
\(
\begin{aligned}
& \therefore \quad b=(n+2)^{\text {th }} \text { term } \\
& \Rightarrow \quad b=a+(n+1) d \\
& \Rightarrow \quad d=\frac{b-a}{n+1}
\end{aligned}
\)
Now,
\(
A_1=a+d \Rightarrow A_1=\left(a+\frac{b-a}{n+1}\right)
\)
\(
\begin{aligned}
&\begin{aligned}
& A_2=a+2 d \Rightarrow A_2=\left(a+\frac{2(b-a)}{n+1}\right) \\
& A_n=a+n d \Rightarrow A_n=\left(a+n \frac{(b-a)}{n+1}\right)
\end{aligned}\\
&\text { These are the required arithmetic means between } a \text { and } b \text {. }
\end{aligned}
\)
Remark
Example 24: If \(\frac{a^n+b^n}{a^{n-1}+b^{n-1}}\) is the \(A M\) between \(a\) and \(b\), then the value of \(n\) is [JEE (WB) 2007]
(a) 0
(b) 1
(c) -1
(d) none of these
Solution: (b) Since \(\frac{a^n+b^n}{a^{n-1}+b^{n-1}}\) is the AM of \(a\) and \(b\).
\(
\begin{aligned}
& \therefore \quad \frac{a^n+b^n}{a^{n-1}+b^{n-1}}=\frac{a+b}{2} \\
& \Rightarrow \quad 2 a^n+2 b^n=a^n+b^n+a b^{n-1}+a^{n-1} b \\
& \Rightarrow \quad a^{n-1}(a-b)=b^{n-1}(a-b) \\
& \Rightarrow \quad\left(\frac{a}{b}\right)^{n-1}=1 \Rightarrow n-1=0 \quad \Rightarrow n=1
\end{aligned}
\)
ALITER: We have,
\(
\begin{aligned}
& \frac{a^n+b^n}{a^{n-1}+b^{n-1}}=\frac{a+b}{2} \\
\Rightarrow \quad & \frac{a^n+b^n}{a^{n-1}+b^{n-1}}=\frac{a+b}{2}=\frac{a^1+b^1}{a^0+b^0} \Rightarrow n=1
\end{aligned}
\)
Example 25: The arithmetic mean between two numbers is \(A\), and \(S\) is the sum of \(n\) arithmetic means between these numbers, then
(a) \(S=n A\)
(b) \(A=n S\)
(c) \(A=S\)
(d) none of these
Solution: (a) Let \(x\) and \(y\) be two numbers.
We have, \(A=\frac{x+y}{2} \text { and } S=n\left(\frac{x+y}{2}\right) \Rightarrow S=n A\)
An Important Property of A.M.’s
The sum of \(n\) arithmetic means between two numbers is \(n\) times the single A.M. between them.
Proof: Let \({A}_1, A_2, \ldots, A_n\) be \(n\) arithmetic means between \(a\) and \(b\). Then, \(a, A_1, A_2, \ldots, A_n, b\) is an A.P. with common difference \(d=\frac{\boldsymbol{b}-\boldsymbol{a}}{\boldsymbol{n}+1}\).
Now,
\(
\begin{aligned}
A_1+ & A_2+\cdots+A_n \\
= & \frac{n}{2}\left[A_1+A_n\right] \\
= & \frac{n}{2}[a+b] \\
& {\left[\because a, A_1, A_2, \ldots, A_n, b \text { is an A.P., } \therefore a+b=A_1+A_n\right] } \\
= & n\left(\frac{a+b}{2}\right) \\
= & n \times(\text { A.M. between } a \text { and } b)
\end{aligned}
\)
Example 26: Insert three arithmetic means between 3 and 19.
Solution: Let \(A_1, A_2\), and \(A_3\) be three A.M.’s between 3 and 19 . Then \(3, A_1, A_2, A_3, 19\) are in A.P. whose common difference is
\(
\begin{aligned}
&\begin{array}{ll}
& d=\frac{19-3}{3+1}=4 \\
\therefore \quad & A_1=a+d=3+d=7 \\
\Rightarrow \quad & A_2=a+2d=3+2 d=11 \\
& A_3=a+3d=3+3 d=15
\end{array}\\
&\text { Hence, the required A.M.’s are } 7,11,15 \text {. }
\end{aligned}
\)
Example 27: If eleven A.M.’s are inserted between 28 and 10, then find the number of integral A.M.’s.
Solution: Assume \(A_1, A_2, A_3, \ldots, A_{11}\) be the eleven A.M.’s between 28 and 10 , so \(28, A_1, A_2, \ldots, A_{11}, 10\) are in A.P. Let \(d\) be the common difference of the A.P. The number of terms is 13. Now,
\(
\begin{aligned}
& 10=T_{13}=T_1+12 d=28+12 d \\
\Rightarrow & d=\frac{10-28}{12}=-\frac{18}{12}=-\frac{3}{2}
\end{aligned}
\)
Here integral A.M.’s are
Since your common difference is \(d=-\frac{3}{2}\), the \(k^{\text {th }}\) mean is calculated as:
\(
A_k=28+k \cdot d=28-k\left(\frac{3}{2}\right)
\)
For \(A_k\) to be an integer, the term \(k\left(\frac{3}{2}\right)\) must be an integer. This only happens when \(k\) is an even number (a multiple of 2), because that is the only way to cancel out the 2 in the denominator.
Identifying the Integral Means
Since we inserted 11 means, \(k\) can be any integer from 1 to 11. The even values for \(k\) are:
\(k=2 \Longrightarrow A_2=28-2\left(\frac{3}{2}\right)=25\)
\(k=4 \Longrightarrow A_4=28-4\left(\frac{3}{2}\right)=22\)
\(k=6 \Longrightarrow A_6=28-6\left(\frac{3}{2}\right)=19\)
\(k=8 \Longrightarrow A_8=28-8\left(\frac{3}{2}\right)=16\)
\(k=10 \Longrightarrow A_{10}=28-10\left(\frac{3}{2}\right)=13\)
Total count = 5.
Example 28: Between 1 and 31 are inserted \(m\) arithmetic means so that the ratio of the \(7^{\text {th }}\) and ( \(\left.m-1\right)^{\text {th }}\) means is \(5: 9\). Find the value of \(m\).
Solution: Let \(A_1, A_2, \ldots, A_m\) be \(m\) arithmetic means between 1 and 31. Then, \(1, A_1, A_2, \ldots, A_m, 31\) is an A.P. with common difference
\(
d=\frac{31-1}{m+1}=\frac{30}{m+1} \quad\left[\text { Using } d=\frac{b-a}{n+1}\right]
\)
Now,
\(
\begin{aligned}
A_7 & =1+7 d \\
\Rightarrow \quad A_7 & =1+\frac{7 \times 30}{m+1}=\frac{m+211}{m+1} \\
A_{m-1} & =1+(m-1) d \\
& =1+\frac{30}{m+1}(m-1) \\
& =\frac{31 m-29}{m+1}
\end{aligned}
\)
It is given that
\(
\begin{aligned}
& \frac{A_7}{A_{m-1}}=\frac{5}{9} \\
\Rightarrow & \frac{m+211}{31 m-29}=\frac{5}{9} \\
\Rightarrow & 9 m+1899=155 m-145 \\
\Rightarrow & 146 m=2044 \\
\Rightarrow & m=14
\end{aligned}
\)
Example 29: For what value of \(n,\left(a^{n+1}+b^{n+1}\right) /\left(a^n+b^n\right)\) is the arithmetic mean of \(\boldsymbol{a}\) and \(\boldsymbol{b}\) ?
Solution: Since A.M. of \(a\) and \(b\) is \((a+b) / 2\), we have
\(
\begin{aligned}
& \frac{a^{n+1}+b^{n+1}}{a^n+b^n}=\frac{a+b}{2} \\
\Rightarrow & 2\left(a^{n+1}+b^{n+1}\right)=\left(a^n+b^n\right)(a+b) \\
\Rightarrow & 2 a^{n+1}+2 b^{n+1}=a^{n+1}+a^n b+b^n a+b^{n+1} \\
\Rightarrow & a^{n+1}+b^{n+1}=a^n b+b^n a \\
\Rightarrow & a^n(a-b)=b^n(a-b) \\
\Rightarrow & a^n=b^n \\
\Rightarrow & \frac{a^n}{b^n}=1 \\
\Rightarrow & \left(\frac{a}{b}\right)^n=1 \\
\Rightarrow & \left(\frac{a}{b}\right)^n=\left(\frac{a}{b}\right)^0 \\
\Rightarrow & n=0
\end{aligned}
\)
Geometric Progression (G.P.)
G.P. is a sequence of numbers whose first term is non-zero and each of the succeeding terms is equal to the proceeding terms multiplied by a constant. Thus in a G.P., the ratio of successive terms is constant. This constant factor is called the common ratio of the series and is obtained by dividing any term by that which immediately proceeds it.
Definition: A sequence of non-zero numbers is called a geometric progression (abbreviated as G.P.) if the ratio of a term and the term preceding to it is always a constant quantity.
If \(a\) is the first term and \(r\) is the common ratio, then G.P. can be written as \(a, a r, a r^2, a r^3, a r^4, \ldots, a r^{n-1}\).
\(n^{\text {th }}\) term: \(T_n=a r^{n-1}=l\) (last term), where \(r=\frac{T_n}{T_{n-1}}\).
\(n^{\text {th }}\) term from end: \(T_n^{\prime}=\frac{l}{r^{n-1}}\), where \(l\) is the last term.
Key Characteristics
Definition: A sequence of non-zero numbers where the ratio of any term to its preceding term is constant.
First Term: \({a}\) (non-zero).
Common Ratio: \(r\) (the constant factor, also found by \(T_n / T_{n-1}\) ).
General Term ( \(n^{\text {th }}\) Term): \(T_n=a r^{n-1}\).
\(n^{\text {th }}\) Term from the End: \(T_n^{\prime}=\frac{l}{r^{n-1}}\), where \(l\) is the last term.
Example: In \(2,6,18,54, \ldots, a=2\) and \(r=3\).
Formulas:
Example 1: The third term of a G.P. is 4. The product of first five terms is
(a) \(4^3\)
(b) \(4^5\)
(c) \(4^4\)
(d) none of these
Solution: (b) Let \(a\) be the first term and \(r\) be the common ratio.
We have,
\(
a_3=4 \Rightarrow a r^2=4
\)
∴ Product of first five terms \(=a ~a r ~a r^2 ~a r^3 ~a r^4=\left(a r^2\right)^5=4^5\)
Increasing and Decreasing G.P.
For a G.P. to be increasing or decreasing, \(r>0\). Since if \(r<0\), terms of G.P. are alternately positive and negative and so neither increasing nor decreasing.
If \(a>0\), then G.P. is increasing if \(r>1\) and decreasing if \(0< r<1\). If \(a<0\), then G.P. is decreasing if \(r>1\) and increasing if\(0<r<1\). The above discussion can be exhibited as follows:
\(
\begin{array}{|c|c|c|c|c|}
\hline a & a>0 & a>0 & a<0 & a<0 \\
\hline r & r>1 & 0<r<1 & r>1 & 0<r<1 \\
\hline \text { Result } & \text { Increasing } & \text { Decreasing } & \text { Decreasing } & \text { Increasing } \\
\hline
\end{array}
\)
Example 2: Which term of the G.P. \(2,1,1 / 2,1 / 4, \ldots\) is \(1 / 128\) ?
Solution: Clearly, the given progression is a G.P. with first term \(a=2\) and common ratio \(r=1 / 2\). Let the \(n^{\text {th }}\) term be \(1 / 128\). Then,
\(
\begin{aligned}
& a_n=\frac{1}{128} \\
\Rightarrow & a r^{n-1}=\frac{1}{128} \\
\Rightarrow & 2\left(\frac{1}{2}\right)^{n-1}=\frac{1}{128} \\
\Rightarrow & \left(\frac{1}{2}\right)^{n-2}=\left(\frac{1}{2}\right)^7 \\
\Rightarrow & n-2=7 \\
\Rightarrow & n=9
\end{aligned}
\)
Thus, \(9^{\text {th }}\) term of the given G.P. is \(1 / 128\).
Example 3: The first term of a G.P. is 1. The sum of the third and fifth terms is 90 . Find the common ratio of the G.P.
Solution: Let \(r\) be the common ratio of the G.P. It is given that the first term is \(a=1\). Now,
\(
\begin{aligned}
& a_3+a_5=90 \\
\Rightarrow & a r^2+a r^4=90 \\
\Rightarrow & r^2+r^4=90 \\
\Rightarrow & r^4+r^2-90=0
\end{aligned}
\)
\(
\begin{aligned}
& \Rightarrow r^4+10 r^2-9 r^2-90=0 \\
& \Rightarrow\left(r^2+10\right)\left(r^2-9\right)=0 \\
& \Rightarrow r^2-9=0 \\
& \Rightarrow r= \pm 3
\end{aligned}
\)
Example 4: Fifth term of a G.P. is 2. Find the product of its first nine terms.
Solution: \(t_5=a r^4=2\)
Product of its first 9 terms is
\(
\begin{aligned}
a(a r)\left(a r^2\right) \cdots\left(a r^8\right) & =a^9 r^{1+2+\cdots+8} \\
& =a^9 r^{\frac{8}{2}(1+8)} \\
& =a^9 r^{36} \\
& =\left(a r^{{4}}\right)^9=2^9=512
\end{aligned}
\)
Example 5: The fourth, seventh and the last term of a G.P. are 10, 80 and 2560, respectively. Find the first term and the number of terms in the G.P.
Solution: Let \(a\) be the first term and \(r\) be the common ratio of the given G.P. Then,
\(
\begin{aligned}
& a_4=10, a_7=80 \Rightarrow a r^3=10 \text { and } a r^6=80 \\
\Rightarrow & \frac{a r^6}{a r^3}=\frac{80}{10} \Rightarrow r^3=8 \Rightarrow r=2
\end{aligned}
\)
Putting \(r=2\) in \(a r^3=10\), we get \(a=10 / 8\).
Let there be \(n\) terms in the given G.P. Then,
\(
\begin{aligned}
& a_n=2560 \Rightarrow a r^{n-1}=2560 \\
\Rightarrow & \frac{10}{8}\left(2^{n-1}\right)=2560 \\
\Rightarrow & 2^{n-4}=256 \Rightarrow 2^{n-4}=2^8 \\
\Rightarrow & n-4=8 \Rightarrow n=12
\end{aligned}
\)
Example 6: Three numbers are in G.P. If we double the middle term, we get an A.P. Then find the common ratio of the G.P.
Solution: Step 1: Define the terms of the G.P. and A.P.
Let the three numbers in G.P. be \(a\), \(a r\), and \(a r^2\), where \(a\) is the first term and \(r\) is the common ratio.
When the middle term is doubled, the new sequence for the A.P. is \(a, 2 a r\), and \(a r^2\)
Step 2: Apply the condition for an A.P. and form the equation
For three terms \(x, y, z\) to be in A.P., the condition \(2 y=x+z\) must hold.
Applying this condition to the A.P. sequence \(a, 2 a r, a r^2\) :
\(
\begin{gathered}
2(2 a r)=a+a r^2 \\
4 a r=a+a r^2
\end{gathered}
\)
Step 3: Solve the equation for the common ratio \(r\)
Rearrange the equation into a quadratic form:
\(
a r^2-4 a r+a=0
\)
Factor out \(a\) :
\(
a\left(r^2-4 r+1\right)=0
\)
Since the numbers are in a G.P., \(a \neq 0\), so we can divide by \(a\) :
\(
r^2-4 r+1=0
\)
Use the quadratic formula \(r=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\) with \(a=1, b=-4, c=1\) to solve for \(r:\)
\(
\begin{gathered}
r=\frac{4 \pm \sqrt{(-4)^2-4 \cdot 1 \cdot 1}}{2 \cdot 1} \\
r=\frac{4 \pm \sqrt{16-4}}{2} \\
r=\frac{4 \pm \sqrt{12}}{2} \\
r=\frac{4 \pm 2 \sqrt{3}}{2} \\
r=2 \pm \sqrt{3}
\end{gathered}
\)
Example 7: Three numbers are in G.P. whose sum is 70. If the extremes be each multiplied by 4 and the means by 5, they will be in A.P. Find the numbers.
Solution: Let the numbers be \(a, a r\), and \(a r^2\). Then,
\(
a\left(1+r+r^2\right)=70 \dots(1)
\)
It is given that \(4 a, 5 a r\), and \(4 a r^2\) are in A.P. Therefore,
\(
\begin{aligned}
& 2(5 a r)=4 a+4 a r^2 \\
\Rightarrow & 5 r=2+2 r^2 \\
\Rightarrow & 2 r^2-5 r+2=0 \\
\Rightarrow & (2 r-1)(r-2)=0 \\
\Rightarrow & r=2 \text { or } r=1 / 2
\end{aligned}
\)
Putting \(r=2\) in (1), we obtain \(a=10\). Putting \(r=1 / 2\) in (1), we get \(a=40\). Hence, the numbers are \(10,20,40\) or \(40,20,10\).
Example 8: If \(a, b, c\) be respectively the \(p^{\text {th }} q^{\text {th }}\) and \(r^{\text {th }}\) terms of a G.P., then [AIEEE 2002]
\(
\Delta=\left|\begin{array}{ccc}
\log a & \log b & \log c \\
p & q & r \\
1 & 1 & 1
\end{array}\right| \text { equals }
\)
(a) 1
(b) 0
(c) -1
(d) none of these
Solution: (b) Let \(A\) be the first term and \(R\) be the common ratio of the G.P. Then, the formula for the \(n^{\text {th }}\) term of a G.P. is \(T_n=A R^{n-1}\).
\(
\begin{aligned}
& a=T_p=A R^{p-1} \Rightarrow \log a=\log A+(p-1) \log R \dots(i) \\
& b=T_q==A R^{q-1} \Rightarrow \log b=\log A+(q-1) \log R \dots(ii) \\
& c=T_r=A R^{r-1} \Rightarrow \log c=\log A+(r-1) \log R \dots(iii)
\end{aligned}
\)
Multiplying (i), (ii) and (iii) by \((q-r),(r-p)\) and \((p-q)\) respectively and adding, we get
\(
\begin{aligned}
& (q-r) \log a+(r-p) \log b+(p-q) \log c=0 \\
\Rightarrow \quad & \Delta=0
\end{aligned}
\)
Example 9: Let \(a_1, a_2, a_3, \ldots, a_n, \ldots \ldots\) be a GP such that \(\frac{a_4}{a_6}=\frac{1}{4}\) and \(a_2+a_5=216\). Then, \(a_1=\)
(a) 12 or, \(\frac{108}{7}\)
(b) 10
(c) 7 or, \(\frac{54}{7}\)
(d) none of these
Solution: (a) Let \(r\) be the common ratio of the G.P. Then,
\(
\frac{a_4}{a_6}=\frac{1}{4} \Rightarrow \frac{1}{r^2}=\frac{1}{4} \Rightarrow r= \pm 2
\)
Now,
\(
\begin{aligned}
& a_2+a_5=216 \\
\Rightarrow \quad & a_1\left(r+r^4\right)=216 \\
\Rightarrow \quad & 18 a_1=216 \text { or, } 14 a_1=216 \Rightarrow a_1=12, \frac{108}{7}
\end{aligned}
\)
Example 10: If \(a, b, c, d\) and \(p\) are distinct real numbers such that
\(
\left(a^2+b^2+c^2\right) p^2-2 p(a b+b c+c d)+\left(b^2+c^2+d^2\right) \leq 0
\)
then \(a, b, c, d\) are in [CEE (Delhi) 2001, 2006]
(a) AP
(b) GP
(c) HP
(d) \(a b=c d\)
Solution: (b) Let’s expand and group the terms of the given expression:
\(
\left(a^2 p^2-2 a b p+b^2\right)+\left(b^2 p^2-2 b c p+c^2\right)+\left(c^2 p^2-2 c d p+d^2\right) \leq 0
\)
We can recognize each of these groups as a perfect square of the form \((x-y)^2=x^2- 2 x y+y^2\).
Step 2: Simplifying to Sum of Squares
The inequality becomes:
\(
(a p-b)^2+(b p-c)^2+(c p-d)^2 \leq 0
\)
Step 3: Analyzing the Result
In the set of real numbers, the square of any value is always non-negative ( \(\geq 0\) ). Therefore, a sum of squares can only be less than or equal to zero if the sum is exactly zero.
For the sum to be zero, each individual squared term must be zero simultaneously:
\(a p-b=0 \Longrightarrow p=\frac{b}{a}\)
\(b p-c=0 \Longrightarrow p=\frac{c}{b}\)
\(c p-d=0 \Longrightarrow p=\frac{d}{c}\)
Step 4: Conclusion
Since all these ratios are equal to the same value \(p\), we have:
\(
\frac{b}{a}=\frac{c}{b}=\frac{d}{c}=p
\)
This is the fundamental definition of a Geometric Progression (GP), where the ratio between consecutive terms is constant.
Thus, \(a, b, c, d\) are in GP.
Example 11: In a G.P. of positive terms, any term is equal to the sum of the next two terms. Then, the common ratio of the G.P. is [AIEEE 2007]
(a) \(\frac{\sqrt{5}-1}{2}\)
(b) \(\frac{\sqrt{5}+1}{2}\)
(c) \(-\frac{\sqrt{5}+1}{2}\)
(d) \(\frac{1-\sqrt{5}}{2}\)
Solution: (a) Step 1: Set up the Equation
Let the G.P. be \(a, a r, a r^2, a r^3, \ldots\) where \(a>0\) and \(r>0\) (since the terms are positive).
According to the problem, any term is equal to the sum of the next two terms. Let’s take the first term \(\left(T_1\right)\) for simplicity:
\(
T_1=T_2+T_3
\)
Substitute the G.P. formula \(\left(T_n=a r^{n-1}\right)\) :
\(
a=a r+a r^2
\)
Step 2: Simplify the Quadratic
Since \(a\) is the first term of a G.P. of positive terms, \(a \neq 0\). We can divide both sides by \(a\) :
\(
1=r+r^2
\)
Rearranging this into the standard quadratic form \(\left(a x^2+b x+c=0\right)\) :
\(
r^2+r-1=0
\)
Step 3: Solve for \(r\)
Using the quadratic formula \(r=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\) :
\(a=1\), \(b=1\), \(c=-1\)
\(
\begin{gathered}
r=\frac{-1 \pm \sqrt{(1)^2-4(1)(-1)}}{2(1)} \\
r=\frac{-1 \pm \sqrt{1+4}}{2} \\
r=\frac{-1 \pm \sqrt{5}}{2}
\end{gathered}
\)
Step 4: Apply the Constraint
The problem specifies the G.P. consists of positive terms.
If we choose \(r=\frac{-1-\sqrt{5}}{2}\), the ratio is negative, which would cause the terms to alternate between positive and negative.
Therefore, we must choose the positive root:
\(
r=\frac{-1+\sqrt{5}}{2}=\frac{\sqrt{5}-1}{2}
\)
Example 12: The first two terms of a geometric progression add up to 12. The sum of the third and the fourth terms is 48. If the terms of the geometric progression are alternatively positive and negative, then the first term is [AIEEE 2008]
(a) 12
(b) 4
(c) -4
(d) -12
Solution: (d) Step 1: Formulate the Equations
Let the first term be \(a\) and the common ratio be \(r\). The terms are \(a, a r, a r^2, a r^3, \ldots\)
From the first condition: The sum of the first two terms is 12.
\(
\begin{gathered}
a+a r=12 \\
a(1+r)=12 \dots(1)
\end{gathered}
\)
From the second condition: The sum of the third and fourth terms is 48.
\(
\begin{gathered}
a r^2+a r^3=48 \\
a r^2(1+r)=48 \dots(2)
\end{gathered}
\)
Step 2: Solve for the Common Ratio ( \(r\) )
Divide Equation 2 by Equation 1:
\(
\begin{gathered}
\frac{a r^2(1+r)}{a(1+r)}=\frac{48}{12} \\
r^2=4 \\
r= \pm 2
\end{gathered}
\)
Step 3: Apply the “Alternating Signs” Constraint
The problem states that the terms are alternatively positive and negative.
In a G.P., terms alternate signs if and only if the common ratio ( \(r\) ) is negative.
Therefore, we must choose \(r=-2\).
Step 4: Solve for the First Term (a)
Substitute \(r=-2\) back into Equation 1:
\(
a(1+(-2))=12
\)
\(
a=-12
\)
Example 13: If \(a_1, a_2, a_3\left(a_1>0\right)\) are three successive terms of \(a\) G.P. with common ratio \(r\), the value of \(r\) for which \(a_3>4 a_2-3 a_1\) holds is given by
(a) \(1<r<3\)
(b) \(-3<r<-1\)
(c) \(r>3\) or \(r<1\)
(d) none of these
Solution: (c) Step 1: Define the Terms
Let the first term be \(a_1=a\). Since the terms are in a G.P. with common ratio \(r\) :
\(a_1=a\)
\(a_2=a r\)
\(a_3=a r^2\)
We are given that \(a>0\).
Step 2: Set up the Inequality
Substitute these terms into the given condition \(a_3>4 a_2-3 a_1\) :
\(
a r^2>4(a r)-3 a
\)
Step 3: Simplify the Expression
Since \(a>0\), we can divide the entire inequality by \(a\) without reversing the inequality sign:
\(
\begin{gathered}
r^2>4 r-3 \\
r^2-4 r+3>0
\end{gathered}
\)
Step 4: Solve the Quadratic Inequality
Factor the quadratic expression:
\(
(r-1)(r-3)>0
\)
To find the range of \(r\), we identify the roots ( 1 and 3 ) and check the intervals on a number line:
Case \(1(r<1)\) : Both factors are negative, so their product is positive. (Satisfies inequality)
Case \(2(1<r<3)\) : One factor is positive and one is negative, so their product is negative. (Does not satisfy)
Case \(3(r>3)\) : Both factors are positive, so their product is positive. (Satisfies inequality)
Step 5: Conclusion
The inequality holds true when:
\(
r<1 \text { or } r>3
\)
Which corresponds to option (c).
Example 14: If the first and the \(n^{\text {th }}\) terms of \(a\) G.P. are \(a\) and \(b\) respectively and \(P\) is the product of the first \(n\) terms, then \(P^2=\)
(a) \(a b\)
(b) \((a b)^n\)
(c) \((a b)^{n / 2}\)
(d) \((a b)^{2 n}\)
Solution: (b) Method 1: Symmetry Property (The Faster Way)
In any G.P., the product of terms equidistant from the beginning and the end is constant and equal to the product of the first and last terms.
\(T_1 \cdot T_n=a \cdot b\)
\(T_2 \cdot T_{n-1}=(a r) \cdot\left(\frac{b}{r}\right)=a b\)
\(T_3 \cdot T_{n-2}=a b\), and so on.
If we write the product \(P\) twice (once forward and once backward):
\(
\begin{gathered}
P=T_1 \cdot T_2 \cdots T_n \\
P=T_n \cdot T_{n-1} \cdots T_1
\end{gathered}
\)
Multiplying these two equations:
\(
P^2=\left(T_1 T_n\right) \cdot\left(T_2 T_{n-1}\right) \cdots\left(T_n T_1\right)
\)
Since there are \(n\) such pairs, and each pair equals \(a b\) :
\(
\begin{gathered}
P^2=(a b) \cdot(a b) \cdots(a b) \text { (up to } n \text { times) } \\
P^2=(a b)^n
\end{gathered}
\)
Method 2: Let \(r\) be the common ratio of the given G.P. Then,
\(
b=n^{\text {th }} \text { term }=a r^{n-1} \Rightarrow r^{n-1}=\frac{b}{a} \Rightarrow r=\left(\frac{b}{a}\right)^{\frac{1}{n-1}}
\)
and,
\(
\begin{aligned}
& P=\text { Product of the first } n \text { terms } \\
\Rightarrow \quad & P=a \cdot a \cdot a r^2 \ldots a r^{n-1}=a^n r^{1+2+3+\ldots+(n-1)} \\
\Rightarrow \quad & P=a^n r^{\frac{n(n-1)}{2}}=a^n\left\{\left(\frac{b}{a}\right)^{\frac{1}{n-1}}\right\}^{\frac{n(n-1)}{2}}
\end{aligned}
\)
\(
\begin{aligned}
& \Rightarrow \quad P=a^n\left(\frac{b}{a}\right)^{n / 2}=a^{n / 2} b^{n / 2}=(a b)^{n / 2} \\
& \Rightarrow \quad P^2=\left[(a b)^{n / 2}\right]^2=(a b)^n
\end{aligned}
\)
Note: Simplify the exponent of \(r\) Using the sum of the first ( \(n-1\) ) natural numbers, \(\sum= \frac{n(n-1)}{2}\).
Selection of terms of G.P.
Three or more terms in a G.P. may be selected in the following manner:
\(
\begin{array}{clc}
\text { No. of terms } & \text { Terms } & \text { Common ratio } \\
3 & \frac{a}{r}, a, a r & r \\
4 & \frac{a}{r^3}, \frac{a}{r}, a r, a r^3 & r^2 \\
5 & \frac{a}{r^2}, \frac{a}{r}, a, a r, a r^2 & r
\end{array}
\)
Example 15: Three numbers form an increasing G.P. If the middle number is doubled, then the new numbers are in A.P. The common ratio of the G.P. is
(a) \(2-\sqrt{3}\)
(b) \(2+\sqrt{3}\)
(c) \(\sqrt{3}-2\)
(d) \(3+\sqrt{2}\)
Solution: (b) Method-1: Let the three numbers be \(a / r, a, a r\). As the numbers form an increasing G.P. So, \(r>1\). It is given that \(a / r, 2 a, a r\) are in A.P.
\(
\begin{aligned}
& \Rightarrow \quad 4 a=\frac{a}{r}+a r \\
& \Rightarrow \quad r^2-4 r+1=0 \Rightarrow r=2 \pm \sqrt{3} \Rightarrow r=2+\sqrt{3} \quad[\because r>1]
\end{aligned}
\)
Method-2: Step 1: Define the G.P.
Let the three numbers in the increasing G.P. be:
\(
a, a r, a r^2
\)
Since it is an increasing G.P., we know that \(a>0\) and \(r>1\).
Step 2: Form the A.P.
According to the problem, if the middle number ( \(a r\) ) is doubled, the new sequence becomes an Arithmetic Progression (A.P.):
\(
a, 2 a r, a r^2
\)
Step 3: Apply the Property of A.P.
In an A.P., the middle term is the average of the first and third terms (or \(2 \times\) middle term \(=\) first + last).
\(
\begin{gathered}
2(2 a r)=a+a r^2 \\
4 a r=a+a r^2
\end{gathered}
\)
Step 4: Solve for \(r\)
Since \(a\) is a term in an increasing G.P., \(a \neq 0\). We can divide the entire equation by \(a\) :
\(
4 r=1+r^2
\)
Rearranging this into a standard quadratic equation:
\(
r^2-4 r+1=0
\)
Using the quadratic formula \(r=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\) :
\(
r=\frac{4 \pm \sqrt{(-4)^2-4(1)(1)}}{2(1)}=2 \pm \sqrt{3}
\)
Step 5: Select the Correct Value
If \(r=2-\sqrt{3}\), then \(r \approx 2-1.732=0.268\). This would make the G.P. decreasing.
If \(r=2+\sqrt{3}\), then \(r \approx 2+1.732=3.732\). This satisfies the condition of an increasing G.P. ( \(r>1\) ).
Conclusion: The common ratio of the G.P. is \(2+\sqrt{3}\).
Example 16: If the roots of the cubic equation \(a x^3+b x^2+c x+d=0\) are in G.P., then
(a) \(c^3 a=b^3 d\)
(b) \(c a^3=b d^3\)
(c) \(a^3 b=c^3 d\)
(d) \(a b^3=c d^3\)
Solution: (a) Step 1: Define the Roots
Let the roots of the equation \(a x^3+b x^2+c x+d=0\) be in G.P. We can represent them as:
\(
\frac{\alpha}{r}, \alpha, \alpha r
\)
Step 2: Product of Roots
From the relation between roots and coefficients, the product of the roots is given by \(-\frac{d}{a}\) :
\(
\begin{gathered}
\left(\frac{\alpha}{r}\right)(\alpha)(\alpha r)=-\frac{d}{a} \\
\alpha^3=-\frac{d}{a}
\end{gathered}
\)
Since the roots are in G.P., we assume \(\alpha \neq 0\), therefore:
\(
b \alpha=-c \Longrightarrow \alpha=-\frac{c}{b}
\)
Step 3: Substitution
Since \(\alpha\) is a root of the equation, it must satisfy \(a \alpha^3+b \alpha^2+c \alpha+d=0\). Substitute \(a \alpha^3= -d\) into the equation:
\(
\begin{gathered}
(-d)+b \alpha^2+c \alpha+d=0 \\
b \alpha^2+c \alpha=0 \\
\alpha(b \alpha+c)=0
\end{gathered}
\)
Step 4: Final Relation
Now, substitute this value of \(\alpha\) back into the cube relation \(\alpha^3=-\frac{d}{a}\) :
\(
\begin{aligned}
\left(-\frac{c}{b}\right)^3 & =-\frac{d}{a} \\
-\frac{c^3}{b^3} & =-\frac{d}{a} \\
\frac{c^3}{b^3} & =\frac{d}{a} \\
c^3 a & =b^3 d
\end{aligned}
\)
Properties of terms of G.P.
Property-1: If each term of a G.P. is multiplied or divided by some fixed non-zero number, then the resulting sequence is also a G.P.
Proof: Why does this happen?
In a G.P., the ratio of any two consecutive terms \(T_n\) and \(T_{n+1}\) is a constant \(r\) :
\(
\frac{T_{n+1}}{T_n}=r
\)
If we multiply every term by \(k\), the new terms are \(k T_n\) and \(k T_{n+1}\). The new ratio is:
\(
\frac{k T_{n+1}}{k T_n}=\frac{T_{n+1}}{T_n}=r
\)
Because the \(k\) cancels out, the ratio between terms doesn’t change, which is why the common ratio stays the same.
Property-2: If \(x_1, x_2, x_3, \ldots\) and \(y_1, y_2, y_3, \ldots\) are two G.P.’s, then \(x_1 y_1\), \(x_2 y_2, x_3 y_3, \ldots\) and \(\frac{x_1}{y_1}, \frac{x_2}{y_2}, \frac{x_3}{y_3}, \cdots\) are also G.P.’s.
The Rule:
If you have two G.P.s:
Sequence \(x_n: x_1, x_1 r_1, x_1 r_1^2, \ldots\) (Common ratio \(=r_1\) )
Sequence \(y_n: y_1, y_1 r_2, y_1 r_2^2, \ldots\) (Common ratio \(=r_2\) )
Then:
The product sequence \(\left(x_n \cdot y_n\right)\) is a G.P. with a new common ratio of \(r_1 \cdot r_2\).
The quotient sequence \(\left(x_n / y_n\right)\) is a G.P. with a new common ratio of \(r_1 / r_2\).
Multiplication of Two G.P.’s:
Let’s take two different sequences:
G.P. \(1\left(x_n\right): 2,4,8,16, \ldots\) (Ratio \(r_1=2\) )
G.P. \(2\left(y_n\right): 5,15,45,135, \ldots\) (Ratio \(r_2=3\) )
Multiply them term-by-term:
\(x_1 y_1=2 \times 5=10\)
\(x_2 y_2=4 \times 15=60\)
\(x_3 y_3=8 \times 45=360\)
Check the result: Is 10, 60, 360, . . . a G.P.?
\(
\frac{60}{10}=6 \quad \text { and } \quad \frac{360}{60}=6
\)
Yes! It is a G.P. with common ratio 6. Note that \(r_1 \times r_2=2 \times 3=6\).
For Example: Division of Two G.P.s
Using the same sequences:
G.P. \(1\left(x_n\right): 2,4,8,16, \ldots\) (Ratio \(r_1=2\) )
G.P. \(2\left(y_n\right): 5,10,20,40, \ldots\left(\right.\) Ratio \(\left.r_2=2\right)\)
Divide them term-by-term ( \(x_n / y_n\) ):
\(x_1 / y_1=2 / 5=0.4\)
\(x_2 / y_2=4 / 10=0.4\)
\(x_3 / y_3=8 / 20=0.4\)
Check the result: The sequence is \(0.4,0.4,0.4, \ldots\) This is a G.P. where the common ratio is 1 (since \(r_1 / r_2=2 / 2=1\) ).
Why does this work?
For any two terms in the product sequence, the ratio of the \((n+1)^{\text {th }}\) term to the \(n^{\text {th }}\) term is:
\(
\frac{x_{n+1} y_{n+1}}{x_n y_n}=\left(\frac{x_{n+1}}{x_n}\right)\left(\frac{y_{n+1}}{y_n}\right)=r_1 \cdot r_2
\)
Since \(r_1\) and \(r_2\) are constants, their product is also a constant, which fulfills the definition of a G.P.
Property 3: If \(x_1, x_2, x_3, \ldots\) is a G.P. of positive terms, then \(\log x_1\), \(\log x_2, \log x_3, \ldots\) is an A.P. and vice versa.
Case-1: The Forward Rule: GP ⟹AP
If a sequence \(x_1, x_2, x_3, \ldots\) is a G.P. with first term \(a\) and common ratio \(r\), its terms are:
\(
a, a r, a r^2, a r^3, \ldots
\)
When we take the logarithm of each term (let’s use base 10 or \(e\) ):
Term 1: \(\log (a)\)
Term 2: \(\log (a r)=\log a+\log r\)
Term 3: \(\log \left(a r^2\right)=\log a+2 \log r\)
Term 4: \(\log \left(a r^3\right)=\log a+3 \log r\)
Observation: Notice that each term is obtained by adding a constant value, \(\log r\), to the previous term. This is the definition of an Arithmetic Progression (A.P.).
New First Term: \(\log a\)
New Common Difference: \(d=\log r\)
Case-2: The Reverse Rule: AP ⟹ GP
If the sequence \(\log x_1, \log x_2, \log x_3, \ldots\) is an A.P. with common difference \(d\), then:
\(
\log x_{n+1}-\log x_n=d
\)
Using the property of \(\operatorname{logarithms}\left(\log A-\log B=\log \frac{A}{B}\right)\) :
\(
\log \left(\frac{x_{n+1}}{x_n}\right)=d
\)
To solve for the ratio, we exponentiate both sides (assuming base 10):
\(
\frac{x_{n+1}}{x_n}=10^d
\)
Observation: Since \(10^d\) is a constant, the ratio between consecutive terms of the original sequence is constant. Therefore, the original sequence \(x_n\) is a G.P. with common ratio \(r=10^d\).
Property 4: Three terms of a G.P. can be taken as \(a / r, a, a r\) and four terms in G.P. as \(a / r^3 a / r, a r, a r^3\). This presentation is useful if the product of terms is involved in the problem. In other problems, terms should be taken as \(a, a r, a r^2, \ldots \ldots\)
Property 5: The reciprocals of the terms of a given G.P. forms a G.P. If you have a G.P. \(a, a r, a r^2, a r^3, \ldots\) with a common ratio \(r\), the sequence of its reciprocals is:
\(
\frac{1}{a}, \frac{1}{a r}, \frac{1}{a r^2}, \frac{1}{a r^3}, \ldots
\)
The resulting sequence is also a G.P.
The new first term is \(\frac{1}{a}\).
The new common ratio is the reciprocal of the original, which is \(\frac{1}{r}\).
For Example, Let’s take the G.P.: \(2,4,8,16, \ldots\)
First term (a): 2
Common ratio ( \(r\) ): 2
Take the reciprocal of each term:
\(\frac{1}{2}\)
\(\frac{1}{4}\)
\(\frac{1}{8}\)
\(\frac{1}{16}\)
Check for G.P.: Find the ratio between consecutive terms:
\(
\begin{aligned}
& \frac{1 / 4}{1 / 2}=\frac{1}{4} \times \frac{2}{1}=\frac{1}{2} \\
& \frac{1 / 8}{1 / 4}=\frac{1}{8} \times \frac{4}{1}=\frac{1}{2}
\end{aligned}
\)
Result: The new sequence is a G.P. with a common ratio of \(\frac{1}{2}\) (which is indeed \(1 / r\)).
Property-6: If each term of a G.P. be raised to the same power, the resulting sequence also forms a G.P.
For Example,
Squaring a G.P. ( \(k=2\) )
Let’s take the G.P.: \(2,4,8,16, \ldots\)
First term (a): 2
Common ratio \((r): 2\)
Now, square each term:
\(2^2=4\)
\(4^2=16\)
\(8^2=64\)
\(16^2=256\)
New Sequence: 4, 16, 64, 256, . . .
Check ratio: \(16 / 4=4\) and \(64 / 16=4\).
Result: It is a G.P. with common ratio 4 (which is \(r^2\), or \(2^2\)).
Why does this work? (The Proof)
To confirm a sequence is a G.P., we check if the ratio of consecutive terms is constant. For the new sequence, the ratio of the \((n+1)^{\text {th }}\) term to the \(n^{\text {th }}\) term is:
\(
\frac{\left(T_{n+1}\right)^k}{\left(T_n\right)^k}=\left(\frac{T_{n+1}}{T_n}\right)^k
\)
Since we know that for the original G.P., \(\frac{T_{n+1}}{T_n}=r\), then:
\(
\left(\frac{T_{n+1}}{T_n}\right)^k=r^k
\)
Because \(r\) and \(k\) are fixed, \(r^k\) is a constant. This proves the new sequence is a G.P.
Property-7: In a finite G.P. the product of the terms equidistant from the beginning and the end is always same and is equal to the product of the first and last term.
For Example,
Let’s take a finite G.P. with 6 terms: \(2,4,8,16,32,64\)
First term ( \(a_1\) ): 2
Last term ( \(a_6\) ): 64
Common ratio \((r): 2\)
Product of extremes (1st and last):
\(
2 \times 64=128
\)
Product of terms 2 nd from beginning and 2 nd from end:
\(
4 \times 32=128
\)
Product of terms 3rd from beginning and 3rd from end:
\(
8 \times 16=128
\)
As you can see, the product remains 128 in every case.
Property-8: Three non-zero numbers \(a, b, c\) are in G.P. iff \(b^2=a c\).
Proof: For three non-zero numbers \(a, b\), and \(c\) :
They are in G.P. if and only if the ratio between the second and first term is equal to the ratio between the third and second term.
Mathematically: \(\frac{b}{a}=\frac{c}{b}\)
Cross-multiplying gives us: \(b^2=a c\) (or \(b=\sqrt{a c}\) )
For Example,
A Basic G.P.
Consider the numbers 3, 6, 12.
\(a=3\)
\(b=6\)
\(c=12\)
Apply the formula:
\(
\begin{gathered}
b^2=6^2=36 \\
a c=3 \times 12=36
\end{gathered}
\)
Since \(b^2=a c\), the numbers are in G.P.
Property-9: If \(a, b, c\) are in G.P., then \(b\) is known as the geometric mean of \(a\) and \(c\)(\(b=\sqrt{a c}\)).
For Example,
Standard Integers
Find the G.M. of 4 and 9.
\(b=\sqrt{4 \times 9}\)
\(b=\sqrt{36}=6\)
The G.P. is: 4, 6,9 (common ratio \(r=1.5\) )
Property-10: If \(a_1, a_2, \ldots, a_n\) are \(n\) non-zero non-negative numbers, then their geometric mean \(G\) is given by \(G=\left(a_1 a_2 \ldots a_n\right)^{1 / n}\).
For Example, Three Numbers ( \(n=3\) )
Find the G.M. of 2, 4, and 8.
Product: \(2 \times 4 \times 8=64\)
\(G=(64)^{1 / 3}\)
\(G=4\) (Check: \(4 \times 4 \times 4=64\). The product is preserved.)
Example 17: If the continued product of three numbers in a G.P. is 216 and the sum of their products in pairs is 156, find the numbers.
Solution: Let the three numbers be \(a / r, a\), and \(a r\). Then, product \(=216\). Hence, \(a / r \times a \times a r=216 \Rightarrow a^3=216 \Rightarrow a=6\) Sum of the products in pairs is 156 . Hence,
\(
\begin{aligned}
& \frac{a}{r} a+a a r+\frac{a}{r} a r=156 \\
\Rightarrow & a^2\left(\frac{1}{r}+r+1\right)=156 \\
\Rightarrow & 36\left(\frac{1+r^2+r}{r}\right)=156 \\
\Rightarrow & 3\left(r^2+r+1\right)=13 r \\
\Rightarrow & 3 r^2-10 r+3=0 \\
\Rightarrow & (3 r-1)(r-3)=0 \\
\Rightarrow & r=\frac{1}{3} \text { or } r=3
\end{aligned}
\)
Hence, putting the values of \(a\) and \(r\), the required numbers are \(18,6,2\) or \(2,6,18\).
Sum of \(n\) terms of G.P.
Example 18: Determine the number of terms in a G.P., if \(a_1=3, a_n=96\) and \(S_n=189\).
Solution: We start with the formula for the \(n^{\text {th }}\) term of a Geometric Progression:
\(
a_n=a_1 \cdot r^{n-1}
\)
Substitution:
The problem provides the values for the first term ( \(a_1\) ) and the last term ( \(a_n\) ):
\(a_1=3\)
\(a_n=96\)
Plugging these into the formula:
\(
96=3 \cdot r^{n-1}
\)
\(
r^{n-1}=32 \dots(1)
\)
Now,
\(
\begin{aligned}
& S_n=189 \\
\Rightarrow & a_1\left(\frac{r^n-1}{r-1}\right)=189 \\
\Rightarrow & 3\left(\frac{\left(r^{n-1}\right) r-1}{r-1}\right)=189 \\
\Rightarrow & 3\left(\frac{32 r-1}{r-1}\right)=189 \quad \text { [Using (1)] } \\
\Rightarrow & 32 r-1=63 r-63 \\
\Rightarrow & 31 r=62 \\
\Rightarrow & r=2
\end{aligned}
\)
Putting \(r=2\) in (1), we get
\(
\begin{aligned}
& 2^{n-1}=32 \\
\Rightarrow & 2^{n-1}=2^5 \\
\Rightarrow & n-1=5 \\
\Rightarrow & n=6
\end{aligned}
\)
Example 19: Prove that the sum to \(n\) terms of the series \(11+103+1005+\cdots\) is \((10 / 9)\left(10^n-1\right)+n^2\).
Solution: Step 1: Identify the General Term ( \(T_n\) )
Observe the structure of the given numbers:
\(T_1=11=10^1+1\)
\(T_2=103=100+3=10^2+3\)
\(T_3=1005=1000+5=10^3+5\)
Looking at the sequence \(1,3,5, \ldots\), these are odd numbers. The \(n^{\text {th }}\) odd number is given by \((2 n-1)\). Thus, the general term of the series is:
\(
T_n=10^n+(2 n-1)
\)
Step 2: Set up the Sum ( \(S_n\) )
The sum of the series is the sum of all terms from 1 to \(n\) :
\(
S_n=\sum_{k=1}^n T_k=\sum_{k=1}^n\left(10^k+2 k-1\right)
\)
We can split this into two separate summations:
\(
S_n=\underbrace{\left(10^1+10^2+10^3+\cdots+10^n\right)}_{\text {Part 1: G.P. }}+\underbrace{(1+3+5+\cdots+(2 n-1))}_{\text {Part 2: A.P. }}
\)
Step 3: Calculate Part 1: The Geometric Progression
The first part is a G.P. where \(a=10\) and \(r=10\). Using the sum formula \(S=\frac{a\left(r^n-1\right)}{r-1}\) :
\(
\text { Sum of G.P. }=\frac{10\left(10^n-1\right)}{10-1}=\frac{10}{9}\left(10^n-1\right)
\)
Step 4: Calculate Part 2: The Arithmetic Progression
The second part is the sum of the first \(n\) odd numbers. This is a known identity, but we can calculate it using the A.P. sum formula \(S=\frac{n}{2}\) (first + last):
\(
\begin{gathered}
\text { Sum of A.P. }=\frac{n}{2}(1+(2 n-1)) \\
\text { Sum of A.P. }=\frac{n}{2}(2 n)=n^2
\end{gathered}
\)
Step 5: Combine the Results
Adding the two parts together:
\(
S_n=\frac{10}{9}\left(10^n-1\right)+n^2
\)
Example 20: Find the sum of the following series: 5 \(+55+555+\cdots\) to \(n\) terms.
Solution: We have,
\(
\begin{aligned}
& 5+55+555+\cdots \text { to } n \text { terms } \\
& =5[1+11+111+\cdots \text { to } n \text { terms }] \\
& =\frac{5}{9}[9+99+999+\cdots \text { to } n \text { terms }] \\
& =\frac{5}{9}\left[(10-1)+\left(10^2-1\right)+\left(10^3-1\right)+\cdots+\left(10^n-1\right)\right] \\
& =\frac{5}{9}\left[\left(10+10^2+10^3+\cdots+10^n\right)-(1+1+1+\cdots n \text { times })\right] \\
& =\frac{5}{9}\left[10 \times \frac{\left(10^n-1\right)}{10-1}-n\right] \\
& =\frac{5}{9}\left[\frac{10}{9}\left(10^n-1\right)-n\right] \\
& =\frac{5}{81}\left[10^{n+1}-10-9 n\right]
\end{aligned}
\)
Important Result
Example 21: If \(p(x)=\left(1+x^2+x^4+\cdots+x^{2 n-2}\right)\left(1+x+x^2\right. \left.+\cdots+x^{n-1}\right)\) is a polynomial in \(x\), then find possible values of \(n\).
Solution: \(p(x)=\left(\frac{1-x^{2 n}}{1-x^2}\right)\left(\frac{1-x}{1-x^n}\right)=\frac{1+x^n}{1+x}\)
As \(p(x)\) is a polynomial, \(x=-1\) must be a zero of \(1+x^n=0\), i.e., \(1+(-1)^n=0\). Hence, \(n\) is odd.
Sum of an Infinite G.P.
The sum of an infinite G.P. with first term ‘ \(a\) ‘ and common ratio \(r(-1<r<1)\) is given by \(S_{\infty}=\frac{a}{1-r}\)
Note: If \(r \geq 1\), then the sum of an infinite G.P. tends to infinity.
Example 22: Find the sum of the following series:
(a) \((\sqrt{2}+1)+1+(\sqrt{2}-1)+\cdots \infty\)
(b) \(\frac{1}{2}+\frac{1}{3^2}+\frac{1}{2^3}+\frac{1}{3^4}+\frac{1}{2^5}+\frac{1}{3^6}+\cdots \infty\)
Solution: (a)The given series is a geometric series with first term \(a=\sqrt{2}+1\) and the common ratio
\(
r=\frac{1}{\sqrt{2}+1}=\frac{\sqrt{2}-1}{(\sqrt{2}+1)(\sqrt{2}-1)}=\sqrt{2}-1
\)
Hence, the sum to infinity is given by
\(
\begin{aligned}
S & =\frac{a}{1-r} \\
& =\frac{\sqrt{2}+1}{1-(\sqrt{2}-1)} \\
& =\frac{\sqrt{2}+1}{2-\sqrt{2}} \\
& =\frac{\sqrt{2}+1}{\sqrt{2}(\sqrt{2}-1)} \\
& =\frac{(\sqrt{2}+1)^2}{\sqrt{2}(\sqrt{2}-1)(\sqrt{2}+1)} \\
& =\frac{3+2 \sqrt{2}}{\sqrt{2}} \\
& =\frac{4+3 \sqrt{2}}{2}
\end{aligned}
\)
(b) We have,
\(
\begin{aligned}
& \frac{1}{2}+\frac{1}{3^2}+\frac{1}{2^3}+\frac{1}{3^4}+\frac{1}{2^5}+\frac{1}{3^6}+\cdots \text { to } \propto \\
& =\left(\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^5}+\cdots\right)+\left(\frac{1}{3^2}+\frac{1}{3^4}+\frac{1}{3^6}+\cdots\right) \\
& =\left(\frac{(1 / 2)}{1-\left(1 / 2^2\right)}\right)+\left(\frac{\left(1 / 3^2\right)}{1-\left(1 / 3^2\right)}\right) \\
& =\frac{2}{3}+\frac{1}{8} \\
& =\frac{19}{24}
\end{aligned}
\)
Example 23: If each term of an infinite G.P. is twice the sum of the terms following it, then find the common ratio of the G.P.
Solution: Step 1: Define the Variables
Let the infinite G.P. be represented as:
\(
a, a r, a r^2, a r^3, \ldots
\)
\(a\) : The first term.
\(r\) : The common ratio ( \(|r|<1\) for the sum to converge).
Step 2: Set Up the Equation
The problem states that each term is twice the sum of the terms following it. Let’s take the first term ( \(a\) ) as our example:
The term: \(a\)
The terms following it: \(a r, a r^2, a r^3, \ldots\)
The sum of following terms: This is itself an infinite G.P. with first term ar and common ratio \(r\).
The formula for the sum of an infinite G.P. is \(S_{\infty}=\frac{\text { first term }}{1-r}\). Therefore:
\(
\text { Sum of following terms }=\frac{a r}{1-r}
\)
According to the problem:
\(
a=2\left(\frac{a r}{1-r}\right)
\)
Step 3: Solve for \(r\)
We can now simplify the equation to find the value of the common ratio:
Divide both sides by \(a\) (assuming \(a \neq 0\) ):
\(
1=\frac{2 r}{1-r}
\)
Multiply both sides by \((1-r)\) :
\(
1-r=2 r
\)
Add \(r\) to both sides:
\(
1=3 r
\)
Step 4: Solve for \(r\) :
\(
r=\frac{1}{3}
\)
Example 24: Prove that \(6^{1 / 2} \times 6^{1 / 4} \times 6^{1 / 8} \cdots \infty=6\).
Solution: Step 1: Simplify the Expression
When we multiply terms with the same base, we add their exponents. We can rewrite the expression as:
\(
6\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots \infty\right)
\)
Step 2: Identify the Infinite Series
Let \(S\) be the sum of the exponents:
\(
S=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots \infty
\)
This is an infinite geometric progression (G.P.) where:
First term (a): 1/2
Common ratio \((r): 1 / 2\) (since each term is half of the previous one)
Since the absolute value of the common ratio is less than \(1(|1 / 2|<1)\), the sum converges.
Step 3: Calculate the Sum of the Exponents
Using the sum formula for an infinite G.P., \(S_{\infty}=\frac{a}{1-r}\) :
\(
\begin{gathered}
S=\frac{1 / 2}{1-1 / 2} \\
S=\frac{1 / 2}{1 / 2} \\
S=1
\end{gathered}
\)
Step 4: Substitute Back into the Original Expression
Now, replace the series in the exponent with the sum we just calculated:
\(
6^S=6^1=6
\)
Example 25: Sum of infinite number of terms in G.P. is 20 and sum of their squares is 100. Then find the common ratio of G.P.
Solution: \(a+a r+a r^2+\cdots\) to \(\infty=20\)
\(
\begin{aligned}
\Rightarrow & \frac{a}{1-r}=20 \dots(1)\\
& a^2+a^2 r^2+a^2 r^4+\cdots \text { to } \infty=100
\end{aligned}
\)
\(
\Rightarrow \quad \frac{a^2}{1-r^2}=100 \dots(2)
\)
Squaring (1), we have
\(
\frac{a^2}{(1-r)^2}=400 \dots(3)
\)
Dividing (3) by (2), we get
\(
\begin{aligned}
& \frac{\frac{a^2}{(1-r)^2}}{\frac{a^2}{1-r^2}}=\frac{400}{100} \\
\Rightarrow & \frac{1-r^2}{(1-r)^2}=4 \\
\Rightarrow & \frac{1+r}{1-r}=4 \\
\Rightarrow & 1+r=4-4 r \\
\Rightarrow & 5 r=3 \\
\Rightarrow & r=\frac{3}{5}
\end{aligned}
\)
Example 26: If \(f(x)\) is a function satisfying \(f(x+y)=f(x) f(y)\) for all \(x, y \in N\) such that \(f(1)=3\) and \(\sum_{=1}^n f(x)=120\). Then, the value of \(n\) is [IIT 1992]
(a) 4
(b) 5
(c) 6
(d) none of these
Solution: (a) Step 1: Identify the Function \(f(x)\)
The problem states that \(f(x+y)=f(x) f(y)\). This is a characteristic property of exponential functions.
Given \(f(1)=3\), we can find the subsequent terms:
\(f(2)=f(1+1)=f(1) \cdot f(1)=3 \cdot 3=3^2\)
\(f(3)=f(2+1)=f(2) \cdot f(1)=3^2 \cdot 3=3^3\)
Generally, \(f(x)=3^x\) for any \(x \in \mathbb{N}\).
Step 2: Set Up the Summation
The problem gives the sum of the series as 120:
\(
\sum_{x=1}^n f(x)=f(1)+f(2)+f(3)+\cdots+f(n)=120
\)
Substituting our function \(f(x)=3^x\) :
\(
3^1+3^2+3^3+\cdots+3^n=120
\)
Step 3: Solve for \(n\)
The left side is a finite geometric progression (G.P.) where:
First term (a): 3
Common ratio ( \(r\) ): 3
Number of terms: \(n\)
The formula for the sum of a finite G.P. is \(S_n=\frac{a\left(r^n-1\right)}{r-1}\).
\(
\begin{aligned}
& 120=\frac{3\left(3^n-1\right)}{3-1} \\
& 120=\frac{3\left(3^n-1\right)}{2}
\end{aligned}
\)
Now, let’s isolate \(n\) :
Multiply both sides by 2 : \(240=3\left(3^n-1\right)\)
Divide by 3: \(80=3^n-1\)
Add 1: \(81=3^n\)
Express 81 as a power of \(3: 3^4=3^n\)
Therefore, \(n=4\).
Example 27: Let \(a_n\) be the \(n^{\text {th }}\) term of the G.P. of positive numbers. Let \(\sum_{n=1}^{100} a_{2 n}=\alpha\) and \(\sum_{n=1}^{100} a_{2 n-1}=\beta\), such that \(\alpha \neq \beta\), then the common ratio is [IIT 1992]
(a) \(\alpha / \beta\)
(b) \(\beta / \alpha\)
(c) \(\sqrt{\alpha / \beta}\)
(d) \(\sqrt{\beta / \alpha}\)
Solution: (a) Step 1: Define the Terms of the G.P.
Let the G.P. be \(a, a r, a r^2, a r^3, \ldots, a r^n\), where \(a\) is the first term and \(r\) is the common ratio.
Step 2: Expand the Given Summations
We are given two sums, \(\alpha\) and \(\beta\). Let’s look at the terms they contain:
For \(\beta\) (the sum of odd-indexed terms):
\(
\beta=\sum_{n=1}^{100} a_{2 n-1}=a_1+a_3+a_5+\cdots+a_{199}
\)
Substituting the G.P. formula \(a_n=a r^{n-1}\) :
\(
\beta=a+a r^2+a r^4+\cdots+a r^{198}
\)
For \(\alpha\) (the sum of even-indexed terms):
\(
\alpha=\sum_{n=1}^{100} a_{2 n}=a_2+a_4+a_6+\cdots+a_{200}
\)
Substituting the G.P. formula:
\(
\alpha=a r+a r^3+a r^5+\cdots+a r^{199}
\)
Step 3: Establish the Relationship
If we look closely at the expression for \(\alpha\), we can factor out \(r\) from every term:
\(
\alpha=r\left(a+a r^2+a r^4+\cdots+a r^{198}\right)
\)
Notice that the expression inside the parentheses is exactly the same as our expression for \(\beta\).
Therefore:
\(
\alpha=r \cdot \beta
\)
Step 4: Solve for \(r\)
To find the common ratio \(r\), we simply divide \(\alpha\) by \(\beta\) :
\(
r=\frac{\alpha}{\beta}
\)
Example 28: An infinite G.P. has first term ‘ \(x\) ‘ and sum ‘ 5 ‘, then \(x\) belongs to [IIT (S) 2004]
(a) \(x<-10\)
(b) \(-10<x<0\)
(c) \(0<x<10\)
(d) \(x>0\)
Solution: (c) Step 1: The Sum Formula
For an infinite G.P. with first term \(a\) and common ratio \(r\), the sum \(S_{\infty}\) is given by:
\(
S_{\infty}=\frac{a}{1-r}
\)
Given in the problem:
First term \(a=x\)
Sum \(S_{\infty}=5\)
Substituting these into the formula:
\(
5=\frac{x}{1-r}
\)
Step 2: Isolate the Common Ratio ( \(r\) )
To find the range of \(x\), we must first express \(r\) in terms of \(x\) :
\(
\begin{gathered}
5(1-r)=x \\
1-r=\frac{x}{5}
\end{gathered}
\)
\(
r=1-\frac{x}{5}
\)
Step 3: Apply the Convergence Condition
For the sum of an infinite G.P. to exist (converge), the common ratio \(r\) must satisfy the following condition:
\(
-1<r<1
\)
(Note: \(r \neq 0\) because if \(r=0\), the sum is just the first term \(x\), which would mean \(x=5\), a single point within the valid range).
Substitute the expression for \(r\) into the inequality:
\(
-1<1-\frac{x}{5}<1
\)
Step 4: Solve the Inequality
Subtract 1 from all parts:
\(
-2<-\frac{x}{5}<0
\)
Multiply the entire inequality by -5. Remember that multiplying by a negative number reverses the inequality signs:
\(
\begin{gathered}
(-2)(-5)>x>(0)(-5) \\
10>x>0
\end{gathered}
\)
Rewriting this in standard form: \(0<x<10\)
Geometric Means (G.M.’s)
Let \(a\) and \(b\) be two given numbers. If \(n\) numbers \(G_1, G_2, \ldots, G_n\) are inserted between \(a\) and \(b\) such that the sequence \(a, G_1, G_2\), \(\ldots, G_n, b\) is a G.P., then the numbers \(G_1, G_2, \ldots, G_n\) are known as \(n\) G.M.’s between \(a\) and \(b\).
If a single G.M. \(G\) is inserted between two given numbers \(a\) and \(b\), then \(G\) is known as the G.M. between \(a\) and \(b\). Thus, \(G\) is the G.M. between \(a\) and \(b\).
\(\Leftrightarrow \quad a, G, b\) are in G.P.
\(
\Leftrightarrow G^2=a b \Leftrightarrow G=\sqrt{a b}
\)
For example, the G.M. between 4 and 9 is given by
\(
G=\sqrt{4 \times 9}=6
\)
The G.M. between -9 and -4 is given by
\(
G=\sqrt{-9 \times-4}=-6
\)
Note: If \(a\) and \(b\) are two numbers of opposite signs, then geometric mean between them does not exist.
Insertion of \(n\) G.M.’s Between Two Given Numbers \(a\) and \(b\)
Let \(G_1, G_2, \ldots, G_n\) be \(n\) G.M. between two given numbers \(a\) and \(b\). Then, \(a, G_1, G_2, \ldots, G_n, b\) is a G.P. consisting of \(n+2\) terms. Let \(r\) be the common ratio of this G.P. Then,
\(
\begin{aligned}
& b=(n+2)^{\text {th }} \text { term }=a r^{n+1} \\
\Rightarrow & r^{n+1}=\frac{b}{a} \\
\Rightarrow & r=(b / a)^{\frac{1}{n+1}} \\
\Rightarrow & G_1=a r=a\left(\frac{b}{a}\right)^{1 /(n+1)}
\end{aligned}
\)
\(
\begin{aligned}
& G_2=a r^2=a\left(\frac{b}{a}\right)^{2 /(n+1)} \\
& G_n=a r^n=a\left(\frac{b}{a}\right)^{n /(n+1)}
\end{aligned}
\)
An Important Property of G.M.’s
If \(n\) G.M.’s are inserted between two quantities, then the product of \(n\) G.M.’s is the \(n^{\text {th }}\) power of the single G.M. between the two quantities.
Proof: Let \(G_1, G_2, G_3, \ldots, G_n\) be \(n\) G.M.’s between two quantities \(a\) and \(b\). Then, \(a, G_1, G_2, \ldots, G_n, b\) is a G.P. Let \(r\) be the common ratio of this G.P. Then, \(r=(b / a)^{1 /(n+1)}\) and \(G_1=a r\), \(G_2=a r^2, G_3=a r^3, \ldots, G_n=a r^n\). Now,
\(
\begin{aligned}
G_1 & G_2 G_3 \cdots G_n=(a r)\left(a r^2\right)\left(a r^3\right) \cdots\left(a r^n\right) \\
& =a^n r^{\frac{n(n+1)}{2}} \\
& =a^n\left[\left(\frac{b}{a}\right)^{\frac{1}{n+1}}\right]^{\frac{n(n+1)}{2}} \\
& =a^n\left(\frac{b}{a}\right)^{n / 2}=a^{n / 2} b^{n / 2} \\
& =(\sqrt{a b})^n \\
& =G^n
\end{aligned}
\)
where \(G=\sqrt{a b}\) is the single G.M. between \(a\) and \(b\).
Example 29: Insert four G.M.’s between 2 and 486.
Solution: Common ratio of the series is given by
\(
r=\left(\frac{b}{a}\right)^{\frac{1}{n+1}}
\)
\(
\begin{aligned}
&\begin{aligned}
& =\left(\frac{486}{2}\right)^{\frac{1}{4+1}} \\
& =(243)^{1 / 5} \\
& r=3
\end{aligned}\\
&\text { Hence, the four G.M.’s are } 6,18,54,162 \text {. }
\end{aligned}
\)
Example 30: Find the product of three geometric. means between 4 and \(1 / 4\).
Solution: Product of \(n\) G.M.’s is \((\sqrt{a b})^n=\left(\sqrt{4 \frac{1}{4}}\right)^3=1\)
Example 31: Find two numbers whose arithmetic mean is \(\mathbf{3 4}\) and the geometric mean is 16.
Solution: Let the two numbers be \(a\) and \(b\). Then,
\(
\begin{aligned}
& \text { A.M. }=34 \Rightarrow \frac{a+b}{2}=34 \Rightarrow a+b=68, \\
& \text { G.M. }=16 \\
\Rightarrow & \sqrt{a b}=16 \Rightarrow a b=256 \dots(1) \\
\therefore & (a-b)^2=(a+b)^2-4 a b \\
\Rightarrow & (a-b)^2=(68)^2-4 \times 256=3600 \\
\Rightarrow & a-b=60 \dots(2)
\end{aligned}
\)
On solving (1) and (2), we get \(a=64\) and \(b=4\). Hence, the required numbers are 64 and 4.
Some Important Properties of Arithmetic and Geometric Means
Property-1: \(A\) and \(G\) are respectively arithmetic and geometric means between two positive numbers \(a\) and \(b\), then \(A>G\).
Proof: We have,
\(
\begin{aligned}
& A=\frac{a+b}{2} \text { and } G=\sqrt{a b} \\
\therefore \quad & A-G=\frac{a+b}{2}-\sqrt{a b}=\frac{a+b-2 \sqrt{a b}}{2}=\frac{1}{2}(\sqrt{a}-\sqrt{b})^2>0 \\
\Rightarrow \quad & A>G
\end{aligned}
\)
For example: An excellent way to understand this property is by using two numbers that have a clear square root for their product. Let’s pick \(a=2\) and \(b=8\).
Step 1:Calculate the Arithmetic Mean ( \(A\) )
The Arithmetic Mean is the simple average of the two numbers:
\(
\begin{gathered}
A=\frac{a+b}{2} \\
A=\frac{2+8}{2}=\frac{10}{2}=5
\end{gathered}
\)
Step 2: Calculate the Geometric Mean ( \(G\) )
The Geometric Mean is the square root of their product:
\(
\begin{gathered}
G=\sqrt{a \times b} \\
G=\sqrt{2 \times 8}=\sqrt{16}=4
\end{gathered}
\)
Step 3: Compare the Results
As the property states:
\(
5>4 \Longrightarrow A>G
\)
Property-2: If \(A\) and \(G\) are respectively arithmetic and geometric means between two positive quantities \(a\) and \(b\), then the quadratic equation having \(a, b\) as its roots is \(x^2-2 A x+G^2=0\).
Proof: We have,
\(
A=\frac{a+b}{2} \text { and } G=\sqrt{a b}
\)
The equation having \(a\) and \(b\) as its roots is
\(
x^2-x(a+b)+a b=0 \text { or, } x^2-2 A x+G^2=0
\)
Example 32: Let two numbers have arithmetic mean 9 and geometric mean 4. Then, these numbers are the roots of the quadratic equation [AIEEE 2004]
(a) \(x^2-18 x-16=0\)
(b) \(x^2-18 x+16=0\)
(c) \(x^2+18 x-16=0\)
(d) \(x^2+18 x+16=0\)
Solution: (b) If \(A\) and \(G\) are the A.M. and G.M. respectively of two numbers, then the numbers are the roots of the equation
\(
x^2-2 A x+G^2=0
\)
Here, \(A=9\) and \(G=4\).
So, the numbers are the roots of the equation \(x^2-18 x+16=0\)
Property-3: If \(A\) and \(G\) be the A.M. and G.M. between two positive numbers, then the numbers are \(A \pm \sqrt{A^2-G^2}\)
Proof: The equation having its roots as the given numbers is
\(
\begin{aligned}
& x^2-2 A x+G^2=0 \\
\Rightarrow \quad & x=\frac{2 A \pm \sqrt{4 A^2-4 G^2}}{2} \Rightarrow x=A \pm \sqrt{A^2-G^2}
\end{aligned}
\)
For example: Standard Integer Results
Let’s find the numbers whose A.M. is 5 and G.M. is 4.
Identify values: \(A=5, G=4\).
Plug into the formula: \(x=A \pm \sqrt{A^2-G^2}\)
Calculate:
\(
\begin{gathered}
x=5 \pm \sqrt{5^2-4^2} \\
x=5 \pm \sqrt{25-16} \\
x=5 \pm \sqrt{9} \\
x=5 \pm 3
\end{gathered}
\)
Results:
\(x_1=5+3=8\)
\(x_2=5-3=2\)
The two numbers are 2 and 8. (Quick check: \(2+8 / 2=5\) and \(\sqrt{2 \times 8}=4\)).
Property-4: If the A.M. and G.M. between two numbers are in the ratio \(m: n\), then the numbers are in the ratio
\(
m+\sqrt{m^2-n^2}: m-\sqrt{m^2-n^2}
\)
Proof: Let the two numbers be \(a\) and \(b\).
Then, the two numbers are
\(
a=A+\sqrt{A^2-G^2} \text { and } b=A-\sqrt{A^2-G^2}
\)
It is given that
\(
\begin{array}{ll}
& A: G=m: n \Rightarrow A=\lambda m \text { and } G=\lambda n \text { for some } \lambda . \\
\therefore \quad & \frac{a}{b}=\frac{\lambda m+\sqrt{\lambda^2 m^2-\lambda^2 n^2}}{\lambda m-\sqrt{\lambda^2 m^2-\lambda^2 n^2}}=\frac{m+\sqrt{m^2-n^2}}{m-\sqrt{m^2-n^2}} \\
\Rightarrow \quad & a: b=\left(m+\sqrt{m^2-n^2}\right):\left(m-\sqrt{m^2-n^2}\right)
\end{array}
\)
Example 33: If A.M. of two numbers is twice their G.M., then the ratio of the greatest number to the smallest number isÂ
(a) \(7-4 \sqrt{3}: 1\)
(b) \(7+4 \sqrt{3}: 1\)
(c) \(21: 1\)
(d) \(5: 1\)
Solution: (b) We have, \(A: G=2: 1\)
\(\therefore \quad\) Numbers are in the ratio \((2+\sqrt{4-1}):(2-\sqrt{4-1})\)
i.e. \(\quad 2+\sqrt{3}: 2-\sqrt{3}\) or, \(7+4 \sqrt{3}: 1\)
Harmonic Progression(H.P.)
A sequence \(a_1, a_2, a_3, \ldots, a_n, \ldots\) of non-zero numbers is called a harmonic progression (H.P.) or a harmonic sequence, if the sequence \(1 / a_1, 1 / a_2, 1 / a_3, \ldots, 1 / a_n, \ldots\) is an arithmetic progression (A.P.).
For example, \(1, \frac{1}{3}, \frac{1}{5}, \frac{1}{7}, \frac{1}{9}, \ldots\) is a H.P., because \(1,3,5,7,9, \ldots\). is an A.P.
The \(n^{\text {th }}\) term of a H.P. is the reciprocal of the \(n^{\text {th }}\) term of the corresponding A.P. Thus, if \(a_1, a_2, a_3, \ldots, a_n, \ldots\) is a H.P. and the common difference of the corresponding A.P. is \(d\)
i.e. \(\quad d=\frac{1}{a_2}-\frac{1}{a_1} \quad (\text { i.e., } d=1 / a_{n+1}-1 / a_n )\),
then \(a_n=\frac{1}{a+(n-1) d}\), where \(a=\frac{1}{a_1}\).
In other words, the \(n^{\text {th }}\) term of a H.P. is the reciprocal of the \(n^{\text {th }}\) term of the corresponding A.P.
Example 1: Let \(a, b, c\) be in A.P. and \(|a|<1,|b|<1\), \(|c|<1\). If \(x=1+a+a^2+\ldots\) to \(\infty, y=1+b+b^2+\ldots\) to \(\infty\) and, \(z=1+c+c^2+\ldots\) to \(\infty\), then \(x, y, z\) are in [AIEEE 2005]
(a) \(A P\)
(b) \(G P\)
(c) \(H P\)
(d) none of these
Solution: (c) Step 1: Simplify the Expressions for \(x, y\), and \(z\)
The given expressions are infinite geometric series. Since \(|a|<1,|b|<1\), and \(|c|<1\), these series converge. Using the sum formula \(S_{\infty}=\frac{1}{1-r}\) :
\(x=1+a+a^2+\cdots=\frac{1}{1-a} \Longrightarrow 1-a=\frac{1}{x} \Longrightarrow a=1-\frac{1}{x}\)
\(y=1+b+b^2+\cdots=\frac{1}{1-b} \Longrightarrow 1-b=\frac{1}{y} \Longrightarrow b=1-\frac{1}{y}\)
\(z=1+c+c^2+\cdots=\frac{1}{1-c} \Longrightarrow 1-c=\frac{1}{z} \Longrightarrow c=1-\frac{1}{z}\)
Step 2: Use the Condition for A.P.
We are told that \(a, b, c\) are in A.P. This means their common difference is constant:
\(
2 b=a+c
\)
Now, substitute the expressions for \(a, b\), and \(c\) in terms of \(x, y\), and \(z\) :
\(
2\left(1-\frac{1}{y}\right)=\left(1-\frac{1}{x}\right)+\left(1-\frac{1}{z}\right)
\)
Step 3: Simplify the Equation
\(
\begin{gathered}
2-\frac{2}{y}=1-\frac{1}{x}+1-\frac{1}{z} \\
2-\frac{2}{y}=2-\frac{1}{x}-\frac{1}{z}
\end{gathered}
\)
Cancel out the 2 from both sides:
\(
\begin{gathered}
-\frac{2}{y}=-\left(\frac{1}{x}+\frac{1}{z}\right) \\
\frac{2}{y}=\frac{1}{x}+\frac{1}{z}
\end{gathered}
\)
Step 4: Conclusion
By definition, if \(\frac{1}{x}, \frac{1}{y}, \frac{1}{z}\) are in A.P. (which is what \(\frac{2}{y}=\frac{1}{x}+\frac{1}{z}\) represents), then \(x, y, z\) are in Harmonic Progression (H.P.).
Example 2: If \(x>1, y>1, z>1\) are in G.P., then \(\frac{1}{1+\ln x}, \frac{1}{1+\ln y}, \frac{1}{1+\ln z}\) are in [IIT 1998]
(a) \(A P\)
(b) \(H P\)
(c) \(G P\)
(d) none of these
Solution: (b) Step 1: Establish the G.P. Relationship
Since \(x, y, z\) are in G.P., we know:
\(
y^2=x z
\)
Taking the natural logarithm (ln) on both sides:
\(
\begin{gathered}
\ln \left(y^2\right)=\ln (x z) \\
2 \ln y=\ln x+\ln z
\end{gathered}
\)
This equation shows that \(\ln x, \ln y, \ln z\) are in Arithmetic Progression (A.P.).
Step 2: Modify the A.P.
A fundamental property of an A.P. is that if you add the same constant to every term, the resulting sequence is still an A.P.
Since \(\ln x, \ln y, \ln z\) are in A.P., adding 1 to each term preserves the progression:
\(
(1+\ln x),(1+\ln y),(1+\ln z) \text { are in A.P. }
\)
Step 3: Determine the Reciprocal Progression
By definition, if a sequence of terms \(a, b, c\) is in A.P., then their reciprocals \(\frac{1}{a}, \frac{1}{b}, \frac{1}{c}\) are in Harmonic Progression (H.P.).
Therefore, the reciprocals of our terms:
\(
\frac{1}{1+\ln x}, \frac{1}{1+\ln y}, \frac{1}{1+\ln z} \text { are in H.P. }
\)
Example 3: If \(a_1, a_2, a_3, \ldots, a_n\) are in H.P. then the expression \(a_1 a_2+a_2 a_3+\ldots+a_{n-1} a_n\) is equal to [AIEEE 2006]
(a) \(n\left(a_1-a_n\right)\)
(b) \((n-1)\left(a_1-a_n\right)\)
(c) \(n a_1 a_n\)
(d) \((n-1) a_1 a_n\)
Solution: (d) Step 1: Convert H.P. to A.P.
By definition, if \(a_1, a_2, \ldots, a_n\) are in H.P., then their reciprocals are in A.P.:
\(
\frac{1}{a_1}, \frac{1}{a_2}, \frac{1}{a_3}, \ldots, \frac{1}{a_n} \text { are in A.P. }
\)
Let \(d\) be the common difference of this A.P. Then:
\(
d=\frac{1}{a_2}-\frac{1}{a_1}=\frac{a_1-a_2}{a_1 a_2} \Longrightarrow a_1 a_2=\frac{a_1-a_2}{d}
\)
Similarly:
\(
\begin{gathered}
a_2 a_3=\frac{a_2-a_3}{d} \\
a_{n-1} a_n=\frac{a_{n-1}-a_n}{d}
\end{gathered}
\)
Step 2: Sum the Terms
Now, let’s add these expressions together as requested:
\(
\begin{gathered}
S=a_1 a_2+a_2 a_3+\cdots+a_{n-1} a_n \\
S=\frac{1}{d}\left[\left(a_1-a_2\right)+\left(a_2-a_3\right)+\cdots+\left(a_{n-1}-a_n\right)\right]
\end{gathered}
\)
In this “telescoping” sum, all middle terms cancel out:
\(
S=\frac{1}{d}\left(a_1-a_n\right)
\)
Step 3: Find the Value of \(d\)
We know the \(n^{\text {th }}\) term of an A.P. is given by \(T_n=T_1+(n-1) d\). Applying this to our reciprocal A.P.:
\(
\begin{gathered}
\frac{1}{a_n}=\frac{1}{a_1}+(n-1) d \\
(n-1) d=\frac{1}{a_n}-\frac{1}{a_1}=\frac{a_1-a_n}{a_1 a_n}
\end{gathered}
\)
Rearranging for \(1 / d\) :
\(
\frac{1}{d}=\frac{(n-1) a_1 a_n}{a_1-a_n}
\)
Step 4: Substitute back into the Sum
Substitute the value of \(1 / d\) into our simplified sum \(S\) :
\(
S=\left(\frac{(n-1) a_1 a_n}{a_1-a_n}\right) \cdot\left(a_1-a_n\right)
\)
Cancel out the (\(a_1-a_n\)) term:
\(
S=(n-1) a_1 a_n
\)
Example 4: The \(8^{\text {th }}\) and \(14^{\text {th }}\) term of a H.P. are \(1 / 2\) and \(1 / 3\), respectively. Find its \(20^{\text {th }}\) term. Also, find its general term.
Solution: Let the H.P. be \(\frac{1}{a}, \frac{1}{a+d}, \frac{1}{a+2 d}, \cdots, \frac{1}{a+(n-1) d}, \cdots\)
Then,
\(
\begin{aligned}
& a_8=\frac{1}{2} \text { and } a_{14}=\frac{1}{3} \\
\Rightarrow & \frac{1}{a+7 d}=\frac{1}{2} \text { and } \frac{1}{a+13 d}=\frac{1}{3}\left[\because a_n=\frac{1}{a+(n-1) d}\right] \\
\Rightarrow & a+7 d=2 \text { and } a+13 d=3 \\
\Rightarrow & a=\frac{5}{6}, d=\frac{1}{6}
\end{aligned}
\)
Now,
\(
a_{20}=\frac{1}{a+19 d}=\frac{1}{\frac{5}{6}+\frac{19}{6}}=\frac{1}{4}
\)
and
\(
\begin{aligned}
a_n & =\frac{1}{a+(n-1) d} \\
& =\frac{1}{\frac{5}{6}+(n-1) \times \frac{1}{6}} \\
& =\frac{6}{n+4}
\end{aligned}
\)
Example 5: If the \(20^{\text {th }}\) term of a H.P. is 1 and the \(30^{\text {th }}\) term is \(-1 / 17\), then find its largest term.
Solution: Let the H.P. be \(\frac{1}{a}, \frac{1}{a+d}, \frac{1}{a+2 d}, \frac{1}{a+3 d}, \cdots\). We have,
\(
\begin{aligned}
& a_{20}=1 \text { and } a_{30}=-\frac{1}{17} \\
\Rightarrow & \frac{1}{a+19 d}=1 \text { and } \frac{1}{a+29 d}=\frac{-1}{17} \\
\Rightarrow & a+19 d=1 \text { and } a+29 d=-17 \\
\Rightarrow & a=\frac{176}{5}, d=-\frac{9}{5}
\end{aligned}
\)
Let \(n^{\text {th }}\) term be the largest term. We have,
\(
\begin{aligned}
a_n & =\frac{1}{a+(n-1) d} \\
& =\frac{1}{\frac{176}{5}+(n-1)\left(\frac{-9}{5}\right)} \\
& =\frac{5}{176-9(n-1)} \\
& =\frac{5}{185-9 n}
\end{aligned}
\)
Now, \(a_n\) is the largest term if \(185-9 n\) is the least positive integer.
Clearly, \(185-9 n\) attains the least positive integral value, if \(n=20\). Thus, \(20^{\text {th }}\) term of the given H.P. is the largest term which is equal to 1.
Example 6: If first three terms of the sequence 1/16, \(a, b, 1 / 6\) are in geometric series and last three terms are in harmonic series, then find the values of \(a\) and \(b\).
Solution: 1/16, \(a, b\) are in G.P. Hence,
\(
\begin{aligned}
& a^2=\frac{b}{16} \text { or } 16 a^2=b \dots(1) \\
& a, b, \frac{1}{6} \text { are in H.P. Hence, } \\
& b=\frac{2 a \frac{1}{6}}{a+\frac{1}{6}}=\frac{2 a}{6 a+1} \dots(2)
\end{aligned}
\)
From (1) and (2),
\(
\begin{aligned}
& 16 a^2=\frac{2 a}{6 a+1} \\
\Rightarrow & 2 a\left(8 a-\frac{1}{6 a+1}\right)=0 \\
\Rightarrow & 8 a(6 a+1)-1=0 \\
\Rightarrow & 48 a^2+8 a-1=0 \quad(\because a \neq 0) \\
\Rightarrow & (4 a+1)(12 a-1)=0 \\
\therefore & a=-\frac{1}{4}, \frac{1}{12}
\end{aligned}
\)
When \(a=-1 / 4\), then from (1),
\(
b=16\left(-\frac{1}{4}\right)^2=1
\)
When \(a=1 / 12\), then from (1),
\(
b=16\left(\frac{1}{12}\right)^2=\frac{1}{9} .
\)
Therefore, \(a=-1 / 4, b=1\) or \(a=1 / 12, b=1 / 9\).
Insertion of \(n\) Harmonic Means
Let \(a\) and \(b\) be two given numbers. If \(n\) numbers \(H_1, H_2, \ldots, H_n\) are inserted between \(a\) and \(b\) such that the sequence a, \(H_1, H_2\), \(H_3, \ldots, H_n, b\) is a H.P., then \(H_1, H_2, \ldots, H_n\) are called \(n\) harmonic means between \(a\) and \(b\). Now, \(a, H_1, H_2, \ldots, H_n, b\) are in H.P. Hence,
\(
\frac{1}{a}, \frac{1}{H_1}, \frac{1}{H_2} ; \ldots, \frac{1}{H_n}, \frac{1}{b} \text { are in A.P. }
\)
Let \(D\) be the common difference of this A.P. Then,
\(
\frac{1}{b}=(n+2)^{\text {th }} \text { term }
\)
\(
\begin{aligned}
& \Rightarrow \quad \frac{1}{b}=\frac{1}{a}+(n+1) D \\
& \Rightarrow D=\frac{a-b}{(n+1) a b}
\end{aligned}
\)
Thus, if \(n\) harmonic means are inserted between two given numbers \(a\) and \(b\), then the common diference of the corresponding A.P. is given by
\(
D=\frac{a-b}{(n+1) a b}
\)
Also,
\(
\frac{1}{H_1}=\frac{1}{a}+D, \frac{1}{H_2}=\frac{1}{a}+2 D, \ldots, \frac{1}{H_n}=\frac{1}{a}+n D
\)
On putting the value of \(D\), we can obtain the values of \(H_1, H_2\), \(\ldots, \boldsymbol{H}_n\).
Example 7: Insert 3 Harmonic Means between 2 and 12. (Inserting Multiple Harmonic Means)
Solution: Step 1: Identify the given values:
\(a=2, b=12\)
Number of means to insert, \(n=3\)
Step 2: Calculate the common difference ( \(D\) ) of the reciprocal A.P.: Using the formula you derived:
\(
\begin{gathered}
D=\frac{a-b}{(n+1) a b} \\
D=\frac{2-12}{(3+1)(2 \times 12)}=\frac{-10}{4 \times 24}=\frac{-10}{96}=-\frac{5}{48}
\end{gathered}
\)
Step 3: Find the reciprocals of the means:
\(\frac{1}{H_1}=\frac{1}{a}+D=\frac{1}{2}-\frac{5}{48}=\frac{24-5}{48}=\frac{19}{48}\)
\(\frac{1}{H_2}=\frac{1}{a}+2 D=\frac{1}{2}-\frac{10}{48}=\frac{24-10}{48}=\frac{14}{48}\)
\(\frac{1}{H_3}=\frac{1}{a}+3 D=\frac{1}{2}-\frac{15}{48}=\frac{24-15}{48}=\frac{9}{48}\)
Step 4: Final Values ( \(H=\) reciprocals):
\(H_1=\frac{48}{19} \approx 2.52\)
\(H_2=\frac{48}{14}=\frac{24}{7} \approx 3.43\)
\(H_3=\frac{48}{9}=\frac{16}{3} \approx 5.33\)
Harmonic Means of Two Given Numbers
If \(a\) and \(b\) are two non-zero numbers, then the harmonic mean of \(a\) and \(b\) is a number \(H\) such that the sequence \(a, H\), and \(b\) is a H.P. Now, \(a, H\), and \(b\) is a H.P. Hence,
\(
\begin{aligned}
& \frac{1}{a}, \frac{1}{H}, \frac{1}{b} \text { is an A.P. } \\
\Rightarrow & \frac{2}{H}=\frac{1}{a}+\frac{1}{b} \\
\Rightarrow & H=\frac{2 a b}{a+b}
\end{aligned}
\)
Thus, the harmonic mean \(H\) between two numbers \(a\) and \(b\) is given by \(H=(2 a b) /(a+b)\).
Example 8: Insert four H.M.’s between 2/3 and 2/13.
Solution: Let \(d\) be the common difference of corresponding A.P. So,
\(
\begin{aligned}
d & =\frac{\frac{13}{2}-\frac{3}{2}}{5}=1 \\
\therefore \quad & \frac{1}{H_1}=\frac{3}{2}+1=\frac{5}{2} \text { or } H_1=\frac{2}{5} \\
\frac{1}{H_2} & =\frac{3}{2}+2=\frac{7}{2} \text { or } H_2=\frac{2}{7} \\
\frac{1}{H_3} & =\frac{3}{2}+3=\frac{9}{2} \text { or } H_3=\frac{2}{9} \\
\frac{1}{H_4} & =\frac{3}{2}+4=\frac{11}{2} \text { or } H_4=\frac{2}{11}
\end{aligned}
\)
Properties of A.M., G.M. and H.M. of Two Positive Real Numbers
Let \(A, G\) and \(H\) be arithmetic, geometric and harmonic means of two positive numbers \(a\) and \(b\). Then,
\(
A=\frac{a+b}{2}, G=\sqrt{a b} \text { and } H=\frac{2 a b}{a+b}
\)
These three means possess the following properties.
Property-1: \(A \geq G \geq H\)
Proof: We have,
\(
\begin{aligned}
& A=\frac{a+b}{2}, G=\sqrt{a b} \text { and } H=\frac{2 a b}{a+b} \\
\Rightarrow & A-G=\frac{a+b}{2}-\sqrt{a b}=\frac{(\sqrt{a}-\sqrt{b})^2}{2} \geq 0
\end{aligned}
\)
\(
\begin{aligned}
\Rightarrow \quad & A \geq G \dots(1) \\
& G-H=\sqrt{a b}-\frac{2 a b}{a+b}
\end{aligned}
\)
\(
\begin{aligned}
&\begin{aligned}
& =\sqrt{a b}\left\{\frac{a+b-2 \sqrt{a b}}{a+b}\right\} \\
& =\frac{\sqrt{a b}}{a+b}(\sqrt{a}-\sqrt{b})^2 \geq 0 \\
\Rightarrow & G \geq H \dots(2)
\end{aligned}\\
&\text { From (i) and (ii), we get } A \geq G \geq H \text {. }
\end{aligned}
\)
Note: The equality holds in the above result only when \(a=b\).
A.M., G.M. and H.M. of \(n\) positive quantities, \(a_1, a_2, a_3, \ldots, a_n\) is given by
\(
\begin{aligned}
& A=\frac{a_1+a_2+a_3+\cdots+a_n}{n} \\
& G=\left(a_1 a_2 a_3 \cdots a_n\right)^{1 / n}
\end{aligned}
\)
and \(H=\frac{n}{\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}+\cdots+\frac{1}{a_n}}\)
\(A \geq G \geq H\) also holds here.
Example 9: Relationship between \(A, G\), and \(H\)
If the Arithmetic Mean (A.M.) of two numbers is 25 and their Harmonic Mean (H.M.) is 9, find their Geometric Mean (G.M.).
Solution: Recall the property: For any two positive numbers, \(G^2=A \times H\).
Substitute the values:
\(
\begin{gathered}
G^2=25 \times 9 \\
G^2=225
\end{gathered}
\)
Solve for \(G\) :
\(
G=\sqrt{225}=15
\)
Property-2: \(A, G\), and \(H\) form a G.P., i.e., \(G^2=A H\).
Proof: We have,
\(
\begin{aligned}
A H & =\frac{a+b}{2} \times \frac{2 a b}{a+b} \\
& =a b=(\sqrt{a b})^2=G^2
\end{aligned}
\)
Hence, \(G^2=A H\).
Example 10: The A.M. and H.M. between two numbers are 27 and 12, respectively, then find their G.M.
Solution: Let \(A, G\) and \(H\) denote, respectively, the A.M., G.M. and H.M. between the two numbers. Then, \(A=27\) and \(H=12\). Since \(A, G\), and \(H\) are in G.P. Therefore,
\(
G^2=A H
\)
\(
\begin{aligned}
& =27 \times 12 \\
\Rightarrow \quad G & =18
\end{aligned}
\)
Property-3: The equation having \(a\) and \(b\) as its roots is \(x^2-2 A x+G^2=0\).
Proof: The equation having \(a\) and \(b\) as its roots is
\(
\begin{aligned}
& x^2-(a+b) x+a b=0 \\
\Rightarrow & x^2-2 A x+G^2=0
\end{aligned}
\)
\(
\left[\because A=\frac{a+b}{2} \text { and } G=\sqrt{a b}\right]
\)
Property-4: If \(A\) and \(G\) be the A.M. and G.M. between two positive numbers, then the numbers are \(A \pm \sqrt{A^2-G^2}\).
Proof: The equation having its roots as the given numbers is
\(
\begin{aligned}
x^2- & 2 A x+G^2=0 \\
\Rightarrow \quad x & =\frac{2 A \pm \sqrt{4 A^2-4 G^2}}{2} \\
& =A \pm \sqrt{A^2+G^2}
\end{aligned}
\)
Property-5: If \(A, G\), and \(H\) are arithmetic, geometric and harmonic means between three given numbers \(a, b\) and \(c\), then the equation having \(a, b\), and \(c\) as its roots is
\(
x^3-3 A x^2+\frac{3 G^3}{H} x-G^3=0
\)
Proof: We have,
\(
\begin{aligned}
& A=\frac{a+b+c}{3}, G=(a b c)^{1 / 3}, \\
& \frac{1}{H}=\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{3} \\
\therefore \quad & a+b+c=3 A, a b c=G^3
\end{aligned}
\)
and
\(
\frac{3 G^3}{H}=a b+b c+c a
\)
The equation having \(a, b\), and \(c\) as its roots is
\(
\begin{aligned}
& x^3-(a+b+c) x^2+(a b+b c+c a) x-a b c=0 \\
\Rightarrow & x^3-3 A x^2+\frac{3 G^3}{H} x-G^3=0
\end{aligned}
\)
Example 11: If the A.M. between two numbers exceeds their G.M. by 2 and the G.M. exceeds their H.M. by \(8 / 5\), find the numbers.
Solution: \(A-G=2 \dots(1)\)
\(
G-H=8 / 5 \dots(2)
\)
\(G^2=A H\)
\(
=(G+2)(G-8 / 5)
\)
\(
\Rightarrow \quad G=8
\)
\(
\Rightarrow \quad a b=64 \dots(3)
\)
From (1),
\(
\begin{aligned}
& A=10 \\
\Rightarrow \quad & a+b=20 \dots(4)
\end{aligned}
\)
Solving (3) and (4), we get \(a=4\) and \(b=16\) or \(a=16\) and \(b=4\).
Arithmetic and Geometric Sequences
If \(a_1, a_2, a_3, \ldots, a_n \ldots\) is an A.P. and \(b_1, b_2, b_3, \ldots, b_n, \ldots\) is a G.P., then the sequence \(a_1 b_1, a_2 b_2, a_3 b_3, \ldots, a_n, b_n, \ldots\) is said to be an arithmetico-geometric sequence.
For example, \(a,(a+d) r,(a+2 d) r^2,(a+3 d) r^3, \ldots\). is an arith-metico-geometric sequence.
The general term of an arithmetico-geometric sequence is the product of \(n^{\text {th }}\) terms of the corresponding A.P. and G.P.
For example, \(n^{\text {th }}\) term of \(a,(a+d) r,(a+2 d) r^2,(a+3 d) r^3, \ldots \ldots\) is
\(
\left\{a+(n-1) d \mid r^{n-1}\right.
\)
The sum of \(n\) terms of an arithmetico-geometric sequence \(a,(a+d) r,(a+2 d) r^2,(a+3 d) r^3, \ldots\). is given by
\(
S_n= \begin{cases}\frac{a}{1-r}+d r\left(\frac{1-r^{n-1}}{1-r}\right)-\frac{\{a+(n-1) d\} r^n}{1-r} & , \text { when } r \neq 1 \\ \frac{n}{2}\{2 a+(n-1) d\} & , \text { when } r=1\end{cases}
\)
If \(|r|<1\), then \(r^n, r^{n-1} \rightarrow 0\) as \(n \rightarrow \infty\). So, the sum of an infinite arithmetico-geometric sequence is given by
\(
S_{\infty}=\frac{a}{1-r}+\frac{d r}{(1-r)^2}
\)
Example 12: The sum to infinity of the series \(1+2\left(1-\frac{1}{n}\right)+3\left(1-\frac{1}{n}\right)^2+ \dots\) is
(a) \(n^2\)
(b) \(n(n+1)\)
(c) \(n\left(1+\frac{1}{n}\right)^2\)
(d) none of these
Solution: (a) To find the sum to infinity of the series \(1+2\left(1-\frac{1}{n}\right)+3\left(1-\frac{1}{n}\right)^2+\ldots\), we can identify this as an Arithmetico-Geometric Series (AGS).
Step 1: Analyze the Series Structure
Let \(x=1-\frac{1}{n}\). The series can be written in terms of \(x\) as:
\(
S=1+2 x+3 x^2+4 x^3+\ldots
\)
This is a special case of a power series where the coefficients ( \(1,2,3, \ldots\) ) form an Arithmetic Progression and the variables \(\left(1, x, x^2, \ldots\right)\) form a Geometric Progression.
Step 2: Summing the Series
For a series of the form \(S=\sum_{k=1}^{\infty} k x^{k-1}\), the sum formula for \(|x|<1\) is:
\(
S=\frac{1}{(1-x)^2}
\)
Why? This is the derivative of the standard infinite geometric series: Since \(\frac{1}{1-x}= 1+x+x^2+x^3+\ldots\), differentiating both sides with respect to \(x\) gives:
\(
\frac{d}{d x}\left(\frac{1}{1-x}\right)=0+1+2 x+3 x^2+\ldots \frac{1}{(1-x)^2}=1+2 x+3 x^2+\ldots
\)
Step 3: Substitute \(x\) Back
Now, we substitute \(x=1-\frac{1}{n}\) back into the sum formula:
Find ( \(1-x\) ):
\(
1-x=1-\left(1-\frac{1}{n}\right)=\frac{1}{n}
\)
Square the result:
\(
(1-x)^2=\left(\frac{1}{n}\right)^2=\frac{1}{n^2}
\)
Calculate the reciprocal ( \(S\) ):
\(
S=\frac{1}{(1-x)^2}=\frac{1}{1 / n^2}=n^2
\)
Example 13: The value of \(2^{1 / 4} \cdot 4^{1 / 8} \cdot 8^{1 / 16} \cdot 16^{1 / 32} \ldots \ldots\), is [ AIEEE 2002]
(a) 1
(b) 2
(c) \(3 / 2\)
(d) \(5 / 2\)
Solution: (b) We have,
\(
\begin{aligned}
& 2^{1 / 4} \cdot 4^{1 / 8} \cdot 8^{1 / 16} \cdot 16^{1 / 32} \ldots \ldots \ldots \infty \\
& =2^{1 / 4+2 / 8+3 / 16+4 / 32+\ldots \ldots \ldots \ldots \ldots \infty}
\end{aligned}
\)
\(
=2^{\frac{1}{2^2}+\frac{2}{2^3}+\frac{3}{2^4}+\frac{4}{2^5}+\ldots \ldots \ldots \ldots \ldots}
\)
\(
=2^{\frac{1}{2^2}\left\{1+\frac{2}{2}+\frac{3}{2^2}+\frac{4}{2^3}\right\}}=2^{\frac{1}{2^2}}\left\{\frac{1}{1-\frac{1}{2}}+\frac{1 \times 1 / 2}{\left(1-\frac{1}{2}\right)^2}\right\}=2^{\frac{1}{2}(2+2)}=2^1=2
\)
Example 14: If \(3+\frac{1}{4}(3+d)+\frac{1}{4^2}(3+2 d)+\ldots\) to \(\infty=8\), then the value of \(d\) is
(a) 9
(b) 5
(c) 1
(d) none of these
Solution: (a) The given series is an arithmetico-geometric series.
We have \(3+\frac{1}{4}(3+d)+\frac{1}{4^2}(3+2 d)+\ldots \text { to } \infty=8\)
\(
\Rightarrow \quad \frac{3}{1-1 / 4}+\frac{d \times 1 / 4}{(1-1 / 4)^2}=8 \quad\left[\text { Using }: S=\frac{a}{1-r}+\frac{d r}{(1-r)^2}\right]
\)
\(
\Rightarrow \quad 4+\frac{4 d}{9}=8 \Rightarrow d=9
\)
Example 15: The sum to infinity of the series \(1+\frac{2}{3}+\frac{6}{3^2}+\frac{10}{3^3}+\frac{14}{3^4}+\ldots .\). is [AIEEE 2009]
(a) 2
(b) 3
(c) 4
(d) 6
Solution: (b) Let \(S\) be the sum of the given series i.e.,
\(
\begin{aligned}
& S=1+\frac{2}{3}+\frac{6}{3^2}+\frac{10}{3^3}+\frac{14}{3^4}+\ldots . \text { to } \infty \\
\Rightarrow & (S-1)=\frac{2}{3}+\frac{6}{3^2}+\frac{10}{3^3}+\frac{14}{3^4}+\ldots \text { to } \infty \dots(i) \\
\Rightarrow & (S-1) \times \frac{1}{3}=\frac{2}{3^2}+\frac{6}{3^3}+\frac{10}{3^4}+\ldots \text { to } \infty \dots(ii)
\end{aligned}
\)
\(
\begin{aligned}
&\text { Subtracting (ii) from (i), we get }\\
&\begin{aligned}
\frac{2}{3}(S-1) & =\frac{2}{3}+\frac{4}{3^2}+\frac{4}{3^3}+\frac{4}{3^4}+\ldots \text { to } \infty \\
\Rightarrow \quad \frac{2}{3}(S-1) & =\frac{2}{3}+\frac{\frac{4}{3^2}}{1-\frac{1}{3}} \Rightarrow \frac{2}{3}(S-1)=\frac{2}{3}+\frac{2}{3} \Rightarrow S=3
\end{aligned}
\end{aligned}
\)
Example 16: The sum of the series \(1+3 x+6 x^2+10 x^3+\ldots \infty\) is
(a) \(\frac{1}{(1-x)^2}\)
(b) \(\frac{1}{1-x}\)
(c) \(\frac{1}{(1+x)^2}\)
(d) \(\frac{1}{(1-x)^3}\)
Solution: (d) Clearly, \(1,3,6,10, \ldots\) is not an A.P. But, successive differences of terms form on A.P.
Let \(S=1+3 x+6 x^2+10 x^3+15 x^4+\ldots \infty\)
\(
\begin{aligned}
& \Rightarrow \quad S-x S=1+2 x+3 x^2+4 x^3+5 x^4+\ldots \infty \dots(i) \\
& \Rightarrow \quad x(S-x S)=x+2 x^2+3 x^3+4 x^4+\ldots \dots(ii)
\end{aligned}
\)
Subtracting (ii) and (i), we get
\(
\begin{aligned}
& S(1-x)^2=1+x+x^2+x^3+\ldots \text { to } \infty \\
\Rightarrow & S(1-x)^2=\frac{1}{1-x} \Rightarrow S=\frac{1}{(1-x)^3}
\end{aligned}
\)
Sum to \(n\) terms of some Special Series
Example 1: The sum of \(n\) terms of the series \(1+(1+3)+(1+3+5)+\ldots \ldots\), is
(a) \(n^2\)
(b) \(\left\{\frac{n(n+1)}{2}\right\}^2\)
(c) \(\frac{n(n+1)(2 n+1)}{6}\)
(d) none of these
Solution: (c) Step 1: Identify the General Term ( \(a_k\) )
Each term in the series is itself a sum of odd numbers:
\(T_1=1\)
\(T_2=1+3\)
\(T_3=1+3+5\)
The \(k^{\text {th }}\) term of the series is the sum of the first \(k\) odd numbers. We know that the sum of the first \(k\) odd numbers is \(k^2\).
\(
T_k=\sum_{i=1}^k(2 i-1)=k^2
\)
Step 2: Find the Sum of \(n\) Terms ( \(S_n\) )
The total sum of the series is the sum of these general terms \(T_k\) from \(k=1\) to \(n\) :
\(
S_n=\sum_{k=1}^n T_k=\sum_{k=1}^n k^2
\)
Using the standard formula for the sum of the squares of the first \(n\) natural numbers:
\(
S_n=\frac{n(n+1)(2 n+1)}{6}
\)
Example 2: The coefficient of \(x^{99}\) in the expansion of \((x-1)(x-2) \ldots \ldots(x-100)\) is
(a) 5050
(b) 5000
(c) -5050
(d) -5000
Solution: (c) Step 1: Understanding the Polynomial Structure
The expression is a product of 100 linear factors:
\(
P(x)=\left(x-a_1\right)\left(x-a_2\right) \ldots\left(x-a_{100}\right)
\)
where \(a_1=1, a_2=2, \ldots, a_{100}=100\).
Step 2: Identifying the Coefficient
The coefficient of the second-highest power ( \(x^{n-1}\), which is \(x^{99}\) here) is always the negative sum of the roots.
\(
\text { Coefficient of } x^{99}=-(1+2+3+\cdots+100)
\)
Step 3: Calculating the Sum
The sum of the first \(n\) natural numbers is given by the formula:
\(
\sum_{k=1}^n k=\frac{n(n+1)}{2}
\)
Plugging in \(n=100\) :
\(
\operatorname{Sum}=\frac{100(100+1)}{2}=50 \times 101=5050
\)
Since the coefficient is the negative of this sum:
\(
\text { Coefficient }=-5050
\)
Example 3: If \(f: R \rightarrow R\) satisfies \(f(x+y)=f(x)+f(y)\) for all \(x, y \in R\) and \(f(1)=7\), then \(\sum_{r=1}^n f(r)\) is [AIEEE 2003]
(a) \(\frac{7 n(n+1)}{2}\)
(b) \(\frac{7 n}{2}\)
(c) \(\frac{7(n+1)}{2}\)
(d) \(7 n(n+1)\)
Solution: (a) This problem involves a linear functional equation. Let’s break down the properties of this function to find the sum.
Step 1: Identify the nature of \(f(x)\)
The equation \(f(x+y)=f(x)+f(y)\) is known as Cauchy’s Functional Equation. For functions defined over real numbers ( \(\mathbb{R}\) ), the solution to this equation is a linear function of the form:
\(
f(x)=c x
\)
where \(c\) is a constant.
Step 2: Determine the constant \(\boldsymbol{c}\)
We are given that \(f(1)=7\). Substituting \(x=1\) into our general form \(f(x)=c x\) :
\(
f(1)=c(1) \Longrightarrow 7=c
\)
Therefore, the function is:
\(
f(x)=7 x
\)
Step 3: Evaluate the Summation
Now we need to find the sum \(\sum_{r=1}^n f(r)\) :
\(
\sum_{r=1}^n f(r)=f(1)+f(2)+f(3)+\cdots+f(n)
\)
Substituting \(f(r)=7 r\) :
\(
\begin{aligned}
\sum_{r=1}^n 7 r & =7(1)+7(2)+7(3)+\cdots+7(n) \\
& =7(1+2+3+\cdots+n)
\end{aligned}
\)
Using the standard formula for the sum of the first \(n\) natural numbers, \(\frac{n(n+1)}{2}\) :
\(
\text { Sum }=7 \times \frac{n(n+1)}{2}=\frac{7 n(n+1)}{2}
\)
Example 4: If \(n\) is odd, the sum of \(n\) terms of the series \(1^2-2^2+3^2-4^2+5^2-6^2+\ldots \ldots\) is
(a) \(\frac{n(n+1)}{2}\)
(b) \(\frac{-n(n+1)}{2}\)
(c) \(\frac{n(n-1)}{2}\)
(d) \(\frac{-n(n-1)}{2}\)
Solution: (a) Structure the Series:
Since \(n\) is odd, the series will end on a positive squared term. Let’s write the sum \(S_n\) as:
\(
S_n=1^2-2^2+3^2-4^2+5^2-\cdots+n^2
\)
Group the Terms:
We can group the terms in pairs, leaving the last term \(\left(n^2\right)\) by itself. Since there are \(n\) terms and \(n\) is odd, there are \(n-1\) terms before the last term. These \(n-1\) terms form \(\frac{n-1}{2}\) pairs.
\(
S_n=\left(1^2-2^2\right)+\left(3^2-4^2\right)+\left(5^2-6^2\right)+\cdots+n^2
\)
\(
S_n=1^2-2^2+3^2-4^2+5^2-6^2+\ldots+\left\{(n-2)^2-(n-1)^2\right\}+n^2
\)
\(
\begin{aligned}
= & (1-2)(1+2)+(3-4)(3+4)+(5-6)(5+6)+\ldots \\
& +\{(n-2)-(n-1)\}\{(n-2)+(n-1)\}+n^2 \\
= & -\left(1+2+3+\ldots+(n-2)+(n-1)+n^2\right. \\
= & -\frac{n(n-1)}{2}+n^2=\frac{n(n+1)}{2}
\end{aligned}
\)
Summation of some Special Series
In this section, we shall discuss some problems for finding the sum of some series of the form
\(
\begin{array}{r}
\frac{1}{a(a+d)}+\frac{1}{(a+d)(a+2 d)}+\frac{1}{(a+2 d)(a+3 d)}+\ldots+\ldots \\
\ldots \ldots \ldots+\frac{1}{(a+(n-2) d)(a+(n-1) d)}
\end{array}
\)
In order to find the sum of a finite number of terms of such series, we write its each term as the difference of two terms as given below
\(
\begin{gathered}
\frac{1}{a(a+d)}=\frac{1}{d}\left(\frac{1}{a}-\frac{1}{a+d}\right), \\
\frac{1}{(a+d)(a+2 d)}=\frac{1}{d}\left(\frac{1}{a+d}-\frac{1}{a+2 d}\right), \\
\frac{1}{(a+2 d)(a+3 d)}=\frac{1}{d}\left(\frac{1}{a+2 d}-\frac{1}{a+3 d}\right) \quad \text { and so on. } \\
\therefore \quad \frac{1}{a(a+d)}+\frac{1}{(a+d)(a+2 d)}+\frac{1}{(a+2 d)(a+3 d)}+\ldots \\
\ldots . .+\frac{1}{(a+(n-2) d)(a+(n-1) d)} \\
=\frac{1}{d}\left\{\left(\frac{1}{a}-\frac{1}{a+d}\right)+\left(\frac{1}{a+d}-\frac{1}{a+2 d}\right)+\left(\frac{1}{a+2 d}-\frac{1}{a+3 d}\right)+\ldots\right. \\
\left.+\left(\frac{1}{a+(n-2) d}-\frac{1}{a+(n-1) d}\right)\right\} \\
=\frac{1}{d}\left\{\frac{1}{a}-\frac{1}{a+(n-1) d}\right\}=\frac{n-1}{a\{a+(n-1) d\}}
\end{gathered}
\)
Example 5: The sum of the series \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\ldots+\frac{1}{n(n+1)}\) is
(a) \(\frac{1}{n+1}\)
(b) \(1-\frac{1}{n+1}\)
(c) \(\frac{1}{n+1}-1\)
(d) \(1+\frac{1}{n+1}\)
Solution: (b) This problem is a classic example of a Telescoping Series. The key to solving it is to split each term into the difference of two simpler fractions.
Step 1: Identify the General Term ( \(T_r\) )
The \(r^{\text {th }}\) term of the series is:
\(
T_r=\frac{1}{r(r+1)}
\)
Using Partial Fraction Decomposition, we can rewrite this as:
\(
\begin{gathered}
T_r=\frac{(r+1)-r}{r(r+1)}=\frac{r+1}{r(r+1)}-\frac{r}{r(r+1)} \\
T_r=\frac{1}{r}-\frac{1}{r+1}
\end{gathered}
\)
Step 2: Expand the Sum
Now, let’s write out the sum \(S_n\) by substituting \(r=1,2,3, \ldots, n\) :
For \(r=1: T_1=\left(1-\frac{1}{2}\right)\)
For \(r=2: T_2=\left(\frac{1}{2}-\frac{1}{3}\right)\)
For \(r=3: T_3=\left(\frac{1}{3}-\frac{1}{4}\right)\)
…
For \(r=n: T_n=\left(\frac{1}{n}-\frac{1}{n+1}\right)\)
Step 3: Observe the Cancellation
When we add these all together, the intermediate terms cancel each other out:
\(
S_n=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\cdots+\left(\frac{1}{n}-\frac{1}{n+1}\right)
\)
After all the cancellations, only the first part of the first term and the last part of the last term remain:
\(
S_n=1-\frac{1}{n+1}
\)
Example 5: The sum to \(n\) terms of the series \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\ldots\) is
(a) \(\frac{1}{2 n+1}\)
(b) \(\frac{2 n}{2 n+1}\)
(c) \(\frac{n}{2 n+1}\)
(d) \(\frac{2 n}{n+1}\)
Solution: (c) Step 1: Identify the General Term ( \(T_r\) )
The terms in the denominator are odd numbers: \(1,3,5, \ldots\) The \(r^{\text {th }}\) term of an odd sequence is \((2 r-1)\). The term following it is \((2 r+1)\). Therefore, the general term is:
\(
T_r=\frac{1}{(2 r-1)(2 r+1)}
\)
Step 2: Partial Fraction Decomposition
To split this fraction, notice that the difference between the two factors in the denominator is \((2 r+1)-(2 r-1)=2\). We multiply and divide by 2 to create this difference in the numerator:
\(
\begin{aligned}
T_r & =\frac{1}{2}\left[\frac{2}{(2 r-1)(2 r+1)}\right] \\
T_r & =\frac{1}{2}\left[\frac{(2 r+1)-(2 r-1)}{(2 r-1)(2 r+1)}\right]
\end{aligned}
\)
\(
T_r=\frac{1}{2}\left[\frac{1}{2 r-1}-\frac{1}{2 r+1}\right]
\)
Step 3: Expand and Cancel
Now, let’s write out the sum \(S_n\) :
\(
S_n=\frac{1}{2}\left[\left(\frac{1}{1}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{7}\right)+\cdots+\left(\frac{1}{2 n-1}-\frac{1}{2 n+1}\right)\right]
\)
Just like before, the internal terms cancel out:
\(
S_n=\frac{1}{2}\left[1-\frac{1}{2 n+1}\right]
\)
Step 4: Simplify the Result
Combine the terms inside the bracket using a common denominator:
\(
\begin{gathered}
S_n=\frac{1}{2}\left[\frac{(2 n+1)-1}{2 n+1}\right] \\
S_n=\frac{1}{2}\left[\frac{2 n}{2 n+1}\right] \\
S_n=\frac{n}{2 n+1}
\end{gathered}
\)
Example 6: If \(t_n=\frac{1}{4}(n+2)(n+3)\) for \(n=1,2,3, \ldots .\), then \(\frac{1}{t_1}+\frac{1}{t_2}+\frac{1}{t_3}+\ldots .+\frac{1}{t_{2003}}=\)
(a) \(\frac{4006}{3006}\)
(b) \(\frac{4003}{3007}\)
(c) \(\frac{4006}{3008}\)
(d) \(\frac{4006}{3009}\)
Solution: (d) To solve this, we need to find the sum of the reciprocals of the given terms \(t_n\). This will lead us back to a telescoping series.
Step 1: Find the General Term of the Sum
We are given:
\(
t_n=\frac{1}{4}(n+2)(n+3)
\)
We need to sum the reciprocals, so let’s find \(T_n=\frac{1}{t_n}\) :
\(
T_n=\frac{4}{(n+2)(n+3)}
\)
Step 2: Apply Partial Fraction Decomposition
As we saw in previous illustrations, we can split a fraction with consecutive factors in the denominator:
\(
\begin{gathered}
T_n=4\left[\frac{(n+3)-(n+2)}{(n+2)(n+3)}\right] \\
T_n=4\left[\frac{1}{n+2}-\frac{1}{n+3}\right]
\end{gathered}
\)
Step 3: Expand the Sum
We need to calculate the sum from \(n=1\) to \(n=2003\) :
\(
S=\sum_{n=1}^{2003} 4\left[\frac{1}{n+2}-\frac{1}{n+3}\right]
\)
Let’s write out the first few and last terms:
For \(n=1: 4\left(\frac{1}{3}-\frac{1}{4}\right)\)
For \(n=2: 4\left(\frac{1}{4}-\frac{1}{5}\right)\)
For \(n=3: 4\left(\frac{1}{5}-\frac{1}{6}\right)\)
…
For \(n=2003: 4\left(\frac{1}{2005}-\frac{1}{2006}\right)\)
Step 4: Cancel Terms
Notice that the \(-\frac{1}{4}\) cancels with the \(+\frac{1}{4}\), the \(-\frac{1}{5}\) with the \(+\frac{1}{5}\), and so on. Only the first part of the first term and the last part of the last term remain:
\(
S=4\left[\frac{1}{3}-\frac{1}{2006}\right]
\)
Step 5: Simplify
Find a common denominator:
\(
\begin{gathered}
S=4\left[\frac{2006-3}{3 \times 2006}\right] \\
S=4\left[\frac{2003}{6018}\right] \\
S=\frac{2 \times 2003}{3009} \\
S=\frac{4006}{3009}
\end{gathered}
\)
Example 7: The limiting value of the sum to \(n\) terms of the series \(\frac{3}{1^2 \cdot 2^2}+\frac{5}{2^2 \cdot 3^2}+\frac{7}{3^2 \cdot 4^2}+\ldots \ldots \ldots\) as \(n \rightarrow \infty\) is
(a) 0
(b) 2
(c) \(\frac{1}{2}\)
(d) 1
Solution: (d) To find the limiting value of this series as \(n \rightarrow \infty\), we first need to determine the general term (\(T_r\)) and see if it can be expressed in a telescoping format.
Step 1: Identify the General Term ( \(T_r\) )
Looking at the series:
Numerators: \(3,5,7, \ldots\) (This is an AP with \(a=3, d=2\) ). The \(r^{\text {th }}\) term is \(2 r+1\).
Denominators: \(1^2 \cdot 2^2, 2^2 \cdot 3^2, 3^2 \cdot 4^2, \ldots\) The \(r^{\text {th }}\) term is \(r^2(r+1)^2\).
So, the general term is:
\(
T_r=\frac{2 r+1}{r^2(r+1)^2}
\)
Step 2: Partial Fraction Decomposition
Notice a relationship between the numerator and the terms in the denominator:
\(
(r+1)^2-r^2=\left(r^2+2 r+1\right)-r^2=2 r+1
\)
Since the difference of the squares in the denominator equals the numerator, we can split the fraction:
\(
\begin{gathered}
T_r=\frac{(r+1)^2-r^2}{r^2(r+1)^2} \\
T_r=\frac{(r+1)^2}{r^2(r+1)^2}-\frac{r^2}{r^2(r+1)^2} \\
T_r=\frac{1}{r^2}-\frac{1}{(r+1)^2}
\end{gathered}
\)
Step 3: Find the Sum of \(n\) Terms ( \(S_n\) )
Now, we expand the sum from \(r=1\) to \(n\) :
\(
S_n=\left(\frac{1}{1^2}-\frac{1}{2^2}\right)+\left(\frac{1}{2^2}-\frac{1}{3^2}\right)+\left(\frac{1}{3^2}-\frac{1}{4^2}\right)+\cdots+\left(\frac{1}{n^2}-\frac{1}{(n+1)^2}\right)
\)
Just like in previous examples, all middle terms cancel out:
\(
S_n=1-\frac{1}{(n+1)^2}
\)
Step 4: Calculate the Limiting Value ( \(n \rightarrow \infty\) )
We are asked for the limit as \(n\) approaches infinity:
\(
\lim _{n \rightarrow \infty} S_n=\lim _{n \rightarrow \infty}\left(1-\frac{1}{(n+1)^2}\right)
\)
As \(n\) becomes infinitely large, the fraction \(\frac{1}{(n+1)^2}\) approaches 0.
\(
\text { Limit }=1-0=1
\)
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