Types of Progression
Progressions are various types but in this chapter we will studying only three special types of progressions which are following :
Arithmetic Progression (A.P.)
Let us recall some formulae and properties studied earlier.
A sequence \(a_1, a_2, a_3, \ldots, a_n, \ldots\) is called arithmetic sequence or arithmetic progression if \(a_{n+1}=a_n+d, n \in N\), where \(a_1\) is called the first term and the constant term \(d\) is called the common difference of the A.P.
Let us consider an A.P. (in its standard form) with first term \(a\) and common difference \(d\), i.e., \(a, a+d, a+2 d, \ldots\)
Then the \(n^{\text {th }}\) term (general term) of the A.P. is \(a_n=a+(n-1) d\).
We can verify the following simple properties of an A.P. :
Let \(a, a+d, a+2 d, \ldots, a+(n-1) d\) be an A.P. Then
\(
l=a+(n-1) d
\)
\(
S _n=\frac{n}{2}[2 a+(n-1) d]
\)
We can also write, \(S _n=\frac{n}{2}[a+l]\)
Algorithm to determine whether a sequence is an AP or not
Example 1: (i) \(1,3,5,7, \ldots\)
(ii) \(\pi, \pi+e^\pi, \pi+2 e^\pi, \ldots\)
(iii) \(a, a-b, a-2 b, a-3 b, \ldots\)
Solution: (i) Here, 2nd term – 1st term \(=3\) rd term -2 nd term \(=\ldots\). \(\Rightarrow 3-1=5-3=\ldots=2\) which is a common difference.
(ii) Here, 2nd term -1 st term \(=3\) rd term -2 nd term \(=\ldots\) \(\Rightarrow\left(\pi+e^\pi\right)-\pi=\left(\pi+2 e^\pi\right)-\left(\pi+e^\pi\right)=\ldots\)
\(=e^\pi\), which is a common difference.
(iii) Here, 2nd term -1 st term \(=3\) rd term -2 nd term \(=\ldots\) \(\Rightarrow(a-b)-a=(a-2 b)-(a-b)=\ldots\)
\(=-b\), which is a common difference.
Example 2: In an A.P. if \(m^{ th }\) term is \(n\) and the \(n^{ th }\) term is \(m\), where \(m \neq n\), find the \(p^{ th} \) term.
Solution:
\(
\begin{array}{ll}
\text { Solution We have } & a_m=a+(m-1) d=n, \dots(1)\\
\text { and } & a_n=a+(n-1) d=m \dots(2)
\end{array}
\)
Solving (1) and (2), we get
\(
(m-n) d=n-m, \text { or } d=-1, \dots(3)
\)
and \(\quad a=n+m-1 \dots(4)\)
Therefore
\(
\begin{aligned}
a_p & =a+(p-1) d \\
& =n+m-1+(p-1)(-1)=n+m-p
\end{aligned}
\)
Hence, the \(p^{\text {th }}\) term is \(n+m-p\).
Example 3: If the sum of \(n\) terms of an A.P. is \(n P +\frac{1}{2} n(n-1) Q\), where \(P\) and \(Q\) are constants, find the common difference.
Solution: Let \(a_1, a_2, \ldots a_n\) be the given A.P. Then
\(
S _n=a_1+a_2+a_3+\ldots+a_{n-1}+a_n=n P +\frac{1}{2} n(n-1) Q
\)
Therefore
\(
S _1=a_1= P , S _2=a_1+a_2=2 P + Q
\)
So that
\(
a_2= S _2- S _1= P + Q
\)
Hence, the common difference is given by \(d=a_2-a_1=( P + Q )- P = Q\).
Example 4: The sum of \(n\) terms of two arithmetic progressions are in the ratio \((3 n+8):(7 n+15)\). Find the ratio of their \(12^{\text {th }}\) terms.
Solution: Let \(a_1, a_2\) and \(d_1, d_2\) be the first terms and common difference of the first and second arithmetic progression, respectively. According to the given condition, we have
\(
\frac{\text { Sum to } n \text { terms of first A.P. }}{\text { Sumto } n \text { terms of second A.P. }}=\frac{3 n+8}{7 n+15}
\)
or \(\frac{\frac{n}{2}\left[2 a_1+(n-1) d_1\right]}{\frac{n}{2}\left[2 a_2+(n-1) d_2\right]}=\frac{3 n+8}{7 n+15}\)
\(
\text { or } \quad \frac{2 a_1+(n-1) d_1}{2 a_2+(n-1) d_2}=\frac{3 n+8}{7 n+15} \dots(1)
\)
\(
\begin{aligned}
& \text { Now } \quad \frac{12^{\text {th }} \text { term of first A.P. }}{12^{\text {th }} \text { term of second A.P }}=\frac{a_1+11 d_1}{a_2+11 d_2} \\
& \frac{2 a_1+22 d_1}{2 a_2+22 d_2}=\frac{3 \times 23+8}{7 \times 23+15} \text { [By putting } n=23 \text { in (1)] } \\
&
\end{aligned}
\)
Therefore \(\quad \frac{a_1+11 d_1}{a_2+11 d_2}=\frac{12^{\text {th }} \text { term of first A.P. }}{12^{\text {th }} \text { term of second A.P. }}=\frac{7}{16}\)
Hence, the required ratio is \(7: 16\).
Example 5: The income of a person is Rs. \(3,00,000\), in the first year and he receives an increase of Rs. 10,000 to his income per year for the next 19 years. Find the total amount, he received in 20 years.
Solution: Here, we have an A.P. with \(a=3,00,000, d=10,000\), and \(n=20\).
Using the sum formula, we get,
\(
S _{20}=\frac{20}{2}[600000+19 \times 10000]=10(790000)=79,00,000 .
\)
Hence, the person received Rs. \(79,00,000\) as the total amount at the end of 20 years.
Example 6: Show that the sequence \(<t_n>\) defined by \(t_n=5 n+4\) is an AP, also find its common difference.
Solution: We have, \(t_n=5 n+4\)
On replacing \(n\) by \((n-1)\), we get
\(
\begin{aligned}
& t_{n-1}=5(n-1)+4 \\
& \Rightarrow \quad t_{n-1}=5 n-1 \\
& \therefore \quad t_n-t_{n-1}=(5 n+4)-(5 n-1)=5 \\
&
\end{aligned}
\)
Clearly, \(t_n-t_{n-1}\) is independent of \(n\) and is equal to 5 . So, the given sequence is an \(AP\) with common difference 5.
Example 7: Show that the sequence \(<t_n>\) defined by \(t_n=3 n^2+2\) is not an AP.
Solution: We have, \(t_n=3 n^2+2\)
On replacing \(n\) by \((n-1)\), we get
\(
\begin{aligned}
& t_{n-1}=3(n-1)^2+2 \\
& \Rightarrow \quad t_{n-1}=3 n^2-6 n+5 \\
& \therefore \quad t_n-t_{n-1}=\left(3 n^2+2\right)-\left(3 n^2-6 n+5\right) \\
& =6 n-3 \\
&
\end{aligned}
\)
Clearly, \(t_n-t_{n-1}\) is not independent of \(n\) and therefore it is not constant. So, the given sequence is not an AP.
Remark
Arithmetic mean
Given two numbers \(a\) and \(b\). We can insert a number A between them so that \(a, A , b\) is an A.P. Such a number \(A\) is called the arithmetic mean (A.M.) of the numbers \(a\) and \(b\). Note that, in this case, we have
\(
A -a=b- A , \quad \text { i.e., } A =\frac{a+b}{2}
\)
We may also interpret the A.M. between two numbers \(a\) and \(b\) as their average \(\frac{a+b}{2}\). For example, the A.M. of two numbers 4 and 16 is 10 . We have, thus constructed an A.P. 4, 10, 16 by inserting a number 10 between 4 and 16. The natural question now arises : Can we insert two or more numbers between given two numbers so that the resulting sequence comes out to be an A.P. ? Observe that two numbers 8 and 12 can be inserted between 4 and 16 so that the resulting sequence \(4,8,12,16\) becomes an A.P.
More generally, given any two numbers \(a\) and \(b\), we can insert as many numbers as we like between them such that the resulting sequence is an A.P.
Let \(A _1, A _2, A _3, \ldots, A _n\) be \(n\) numbers between \(a\) and \(b\) such that \(a, A _1, A _2, A _3, \ldots\), \(A _n, b\) is an A.P.
Here, \(b\) is the \((n+2)^{\text {th }}\) term, i.e., \(b=a+[(n+2)-1] d=a+(n+1) d\).
This gives
\(
d=\frac{b-a}{n+1} .
\)
Thus, \(n\) numbers between \(a\) and \(b\) are as follows:
\(
\begin{aligned}
& A _1=a+d=a+\frac{b-a}{n+1} \\
& A _2=a+2 d=a+\frac{2(b-a)}{n+1} \\
& A _3=a+3 d=a+\frac{3(b-a)}{n+1} \\
& \text {….. ….. ….. } \\
& A _n=a+n d=a+\frac{n(b-a)}{n+1} . \\
&
\end{aligned}
\)
Example 8: Insert 6 numbers between 3 and 24 such that the resulting sequence is an A.P.
Solution: Let \(A _1, A _2, A _3, A _4, A _5\) and \(A _6\) be six numbers between 3 and 24 such that \(3, A _1, A _2, A _3, A _4, A _5, A _6, 24\) are in A.P. Here, \(a=3, b=24, n=8\).
Therefore, \(24=3+(8-1) d\), so that \(d=3\).
Thus
\(
\begin{array}{ll}
A _1=a+d=3+3=6 ; & A _2=a+2 d=3+2 \times 3=9 ; \\
A _3=a+3 d=3+3 \times 3=12 ; & A _4=a+4 d=3+4 \times 3=15 ; \\
A _5=a+5 d=3+5 \times 3=18 ; & A _6=a+6 d=3+6 \times 3=21 .
\end{array}
\)
Hence, six numbers between 3 and 24 are 6, 9, 12, 15, 18 and 21 .
General Term of an AP
Let ‘ \(a\) ‘ be the first term, ‘ \(d\) ‘ be the common difference and ‘ \(l\) ‘ be the last term of an AP having ‘ \(n\) ‘ terms, where \(n \in N\). Then, AP can be written as \(a, a+d, a+2 d, \ldots, l-2 d, l-d, l\)
(Case-i) \(n\)th Term of an AP from Beginning
1 st term from beginning \(=t_1=a=a+(1-1) d\)
2nd term from beginning \(=t_2=a+d=a+(2-1) d\)
3rd term from beginning \(=t_3=a+2 d=a+(3-1) d\)
\( \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \vdots \quad \vdots \quad \vdots \quad \vdots\)
\(n\)th term from beginning \(=t_n=a+(n-1) d, \forall n \in N\)
Hence, \(n\)th term of an AP from beginning
\(
=t_n=a+(n-1) d=l \text { [last term] }
\)
(Case-ii) \(n\)th Term of an AP from End
1 st term from end \(=t_1^{\prime}=l=l-(1-1) d\)
2nd term from end \(=t_2^{\prime}=l-d=l-(2-1) d\)
3rd term from end \(=t_3^{\prime}=l-2 d=l-(3-1) d\)
\(\quad \quad \quad \quad \quad \quad \quad \quad \vdots \quad \vdots \quad \vdots \quad \vdots\)
\(n\)th term from end \(=t_n^{\prime}=l-(n-1) d, \forall n \in N\)
Hence, \(n\)th term of an AP from end \(t_n^{\prime}=1-(n-1) d=a \text { [first term] }\)
Now, it is clear that
\(
\begin{aligned}
& t_n+t_n^{\prime}=a+(n-1) d+l-(n-1) d=a+l \\
& \text { or } \quad t_n+t_n^{\prime}=a+l
\end{aligned}
\)
i.e. In a finite \(AP\), the sum of the terms equidistant from the beginning and end is always same and is equal to the sum of first and last term.
Remark
Example 9: Find first negative term of the sequence \(20,19 \frac{1}{4}, 18 \frac{1}{2}, 17 \frac{3}{4}, \ldots\)
Solution: The given sequence is an \(AP\) in which first term, \(a=20\) and common difference, \(d=-\frac{3}{4}\). Let the \(n\)th term of the given AP be the first negative term. Then,
\(
\begin{aligned}
& t_n<0 \Rightarrow a+(n-1) d<0 \\
\Rightarrow \quad & 20+(n-1)\left(-\frac{3}{4}\right)<0
\end{aligned}
\)
\(
\begin{array}{ll}
\Rightarrow & 80-3 n+3<0 \\
\Rightarrow & n>\frac{83}{3} \text { or } n>27 \frac{2}{3} \\
\Rightarrow & n=28
\end{array}
\)
Thus, 28th term of the given sequence is the first negative term.
Example 10: If the \(m\) th term of an AP is \(\frac{1}{n}\) and the \(n\)th term is \(\frac{1}{m}\), then find \({mn}^{th}\) term of an AP.
Solution: If \(A\) and \(B\) are constants, then \(r\) th term of \(AP\) is
\(t_r=A r+B\)
\(
\begin{aligned}
& \text { Given, } t_m=\frac{1}{n} \Rightarrow A m+B=\frac{1}{n} \dots(i)\\
& \text { and } t_n=\frac{1}{m} \Rightarrow A n+B=\frac{1}{m} \dots(ii)
\end{aligned}
\)
From Eqs. (i) and (ii), we get \(A=\frac{1}{m n}\) and \(B=0\)
\(
m n \text {th term }=t_{m n}=A m n+B=\frac{1}{m n} \cdot m n+0=1
\)
Hence, \(m n\)th term of the given \(AP\) is 1.
Example 11: If \(|x-1|, 3\) and \(\mid x-3\) | are first three terms of an increasing AP, then find the 6th term of on AP.
Solution: Case I: For \(x<1\),
\(
\begin{aligned}
& |x-1|=-(x-1) \\
& \text { and } |x-3|=-(x-3)
\end{aligned}
\)
\(\therefore 1-x, 3\) and \(3-x\) are in AP.
\(
\begin{array}{rlrl}
\Rightarrow & & 6 & =1-x+3-x \\
\Rightarrow & x & =-1
\end{array}
\)
Then, first three terms are \(2,3,4\), which is an increasing AP.
\(\therefore\) 6th term is 7. \([\because d=1]\)
Case II: For \(1<x<3\),
\(
\begin{aligned}
& |x-1|=x-1 \\
& \text { and } |x-3|=-(x-3)=3-x
\end{aligned}
\)
\(\therefore x-1,3\) and \(3-x\) are in AP.
\(
\begin{array}{ll}
\Rightarrow & 6=x-1+3-x \\
\Rightarrow & 6=2 \text { [impossible] }
\end{array}
\)
Case III: For \(x>3,|x-1|=x-1\) and \(|x-3|=x-3\)
\(\therefore x-1,3\) and \(x-3\) are in AP.
\(
\Rightarrow \quad 6=x-1+x-3 \Rightarrow x=5
\)
Then, first three terms are \(4,3,2\), which is a decreasing \(AP\).
Example 12: In the sequence 1, 2, 2, 3, 3, 3, 4, 4, 4, \(4, \ldots\), where \(n\) consecutive terms have the value \(n\), find the 150th term of the sequence.
Solution: Let the 150 th term \(=n\)
Then, \(1+2+3+\ldots+(n-1)<150<1+2+3+\ldots+n\)
\(
\begin{array}{ll}
\Rightarrow & \frac{(n-1) n}{2}<150<\frac{n(n+1)}{2} \\
\Rightarrow & n(n-1)<300<n(n+1)
\end{array}
\)
Taking first two members
\(
\begin{aligned}
& n(n-1)<300 \Rightarrow n^2-n-300<0 \\
\Rightarrow & \left(n-\frac{1}{2}\right)^2<300+\frac{1}{4} \\
\Rightarrow & 0<n<\frac{1}{2}+\frac{\sqrt{1201}}{2} \\
\Rightarrow & 0<n<17.8 \dots(i)
\end{aligned}
\)
and taking last two members,
\(
\begin{aligned}
& n(n+1)>300 \\
& \Rightarrow \quad \left(n+\frac{1}{2}\right)^2>300+\frac{1}{4}
\end{aligned}
\)
\(
\begin{array}{ll}
\therefore & n>-\frac{1}{2}+\frac{\sqrt{1201}}{2} \\
\Rightarrow & n>16.8 \dots(ii)
\end{array}
\)
From Eqs. (i) and (ii), we get
\(
\begin{aligned}
& 16.8<n<17.8 \\
& \Rightarrow \quad n=17 \\
&
\end{aligned}
\)
Example 13: If \(a_1, a_2, a_3, a_4\) and \(a_5\) are in AP with common difference \(\neq 0\), find the value of \(\sum_{i=1}^5 a_i\) when \(a_3=2\).
Solution: \(\because a_1, a_2, a_3, a_4\) and \(a_5\) are in AP, we have
\(
a_1+a_5=a_2+a_4=a_3+a_3 \left[\because t_n+t_n^{\prime}=a+l\right]
\)
\(
a_1+a_5=a_2+a_4=4 \quad\left[\because a_3=2\right]
\)
\(
\begin{aligned}
& a_1+a_2+a_3+a_4+a_5=4+2+4=10 \\
& \Rightarrow \quad \sum_{i=1}^5 a_i=10
\end{aligned}
\)
Sum of a Stated Number of Terms of an Arithmetic Series
More than 200 yr ago, a class of German School Children was asked to find the sum of all integers from 1 to 100 inclusive. One boy in the class, an eight year old named Carl Fredrick Gauss (1777-1855) who later established his reputation as one of the greatest Mathematicians announced the answer almost at once. The teacher overawed at this asked Gauss to explain how he got this answer. Gauss explained that he had added these numbers in pairs as follows
\(
(1+100),(2+99),(3+98), \ldots
\)
There are \(\frac{100}{2}=50\) pairs. The answer can be obtained by multiplying 101 by 50 to get 5050.
Sum of \(n\) Terms of an AP
Let ‘ \(a\) ‘ be the first term, ‘ \(d\) ‘ be the common difference, ‘ \(l\) ‘ be the last term of an AP having \(n\) terms and \(S_n\) be the sum of \(n\) terms, then
\(
S_n=a+(a+d)+(a+2 d)+\ldots+(l-2 d)+(l-d)+l \dots(i)
\)
Reversing the right hand terms
\(
S_n=l+(l-d)+(l-2 d)+\ldots+(a+2 d)+(a+d)+a \dots(ii)
\)
On adding Eqs. (i) and (ii), we get
\(
2 S_n=(a+l)+(a+l)+(a+l)+\ldots +(a+l)+(a+l)+(a+l)
\)
\(
\begin{aligned}
& =(a+l)+(a+l)+\ldots \text { upto } n \text { terms }=n(a+l) \\
\therefore \quad S_n & =\frac{n}{2}(a+l) \dots(iii)
\end{aligned}
\)
Now, if we substitute the value of \(l\) viz., \(l=a+(n-1) d\), in this formula, we get
\(
\begin{aligned}
& S_n=\frac{n}{2}[a+a+(n-1) d]=\frac{n}{2}[2 a+(n-1) d] \\
& \therefore \quad S_n=\frac{n}{2}[2 a+(n-1) d] \\
&
\end{aligned}
\)
If we substitute the value of \(a~\) viz.,
\(
l=a+(n-1) d
\)
or \(a=l-(n-1) d\) in Eq. (iii), then
\(
S_n=\frac{n}{2}[2 l-(n-1) d]
\)
If we substitute the value of \(a+l~ v i z\).,
\(
\begin{aligned}
t_n+t_n^{\prime} & =a+l \text { in Eq. (iii), then } \\
S_n & =\frac{n}{2}\left( t _{ n }+ t ^{\prime}{ }_n\right)
\end{aligned}
\)
Corollary I: Sum of first \(n\) natural numbers
i.e. \(1+2+3+4+\ldots+n\)
Here, \(a=1 \text { and } d=1\)
\(
\begin{aligned}
\therefore \quad S & =\frac{n}{2}[2 \cdot 1+(n-1) \cdot 1] \\
& =\frac{n(n+1)}{2}
\end{aligned}
\)
Corollary II: Sum of first \(n\) odd natural numbers
i.e., \(1+3+5+\ldots\)
\(
\begin{array}{ll}
\text { Here, } & a=1 \\
\text { and } & d=2 \\
\therefore & S=\frac{n}{2}[2 \cdot 1+(n-1) \cdot 2]=n^2
\end{array}
\)
Corollary III: If sum of first \(n\) terms is \(S_n\), then sum of next \(m\) terms is \(S_{m+n}-S_n\).
Example 14: If \(S_n, t_n\) and \(d\) are sum of \(n\) terms, \(n\)th term and common difference of an AP respectively, then prove that
\(
\begin{array}{rlr}
d=t_n-t_{n-1} & {[n \geq 2]} \\
t_n=S_n-S_{n-1} & {[n \geq 2]} \\
d=S_n-2 S_{n-1}+S_{n-2} & {[n \geq 3]}
\end{array}
\)
Proof:
\(
\begin{aligned}
& \because \quad S_n=t_1+t_2+t_3+\ldots+t_{n-1}+t_n \\
& \Rightarrow \quad S_n=S_{n-1}+t_n \\
& \therefore \quad t_n=S_n-S_{n-1} \\
&
\end{aligned}
\)
\(
\begin{aligned}
& \text { but } \quad d=t_n-t_{n-1} \\
& =\left(S_n-S_{n-1}\right)-\left(S_{n-1}-S_{n-2}\right) \\
& \therefore \quad d=S_n-2 S_{n-1}+S_{n-2} \\
&
\end{aligned}
\)
Example 15: A sequence is an AP if and only if the sum of its \(n\) terms is of the form \(A n^2+B n\), where \(A\) and \(B\) are constants independent of \(n\). In this case, the \(n\)th term and common difference of the AP are \(A(2 n-1)+B\) and \(2 A\), respectively.
Proof: As \(S_n=A n^2+B n\)
\(
\begin{aligned}
\therefore \quad S_{n-1} & =A(n-1)^2+B(n-1) \\
\therefore \quad t_n & =S_n-S_{n-1} \\
& =\left(A n^2+B n\right)-\left[A(n-1)^2+B(n-1)\right] \\
& =A\left[n^2-(n-1)^2\right]+B \\
\Rightarrow \quad t_n & =A(2 n-1)+B \\
t_{n-1} & =A[2(n-1)-1]+B \\
& =A(2 n-3)+B
\end{aligned}
\)
Now, \(t_n-t_{n-1}=[A(2 n-1)+B]-[A(2 n-3)+B]\)
\(=2 A \text { [a constant] }\)
Hence, the sequence is an AP.
Conversely, consider an AP with first term \(a\) and common difference \(d\).
Sum of first \(n\) terms \(=\frac{n}{2}[2 a+(n-1) d]\)
\(
=\frac{d n^2}{2}+\left(a-\frac{d}{2}\right) n=A n^2+B n \text {, }
\)
where, \(A=\frac{d}{2}, B=a-\frac{d}{2}\)
Hence, \(S_n=A n^2+B n\), where \(A\) and \(B\) are constants independent of \(n\).
Hence, the converse is true.
Corollary \(\because \quad S_n=A n^2+B n\)
\(
\therefore \quad t_n=A(2 n-1)+B
\)
\(
\left.t_n=A \text { (replacing } n^2 \text { by } 2 n-1\right)+ \text { coefficient of } n
\)
and \(d=2 A\)
i.e. \(d=2\) [coefficient of \(n^2\) ]
\(
\begin{array}{|c|c|c|c|}
\hline & S _n & t _n & d \\
\hline 1 . & 5 n^2+3 n & 5(2 n-1)+3=10 n-2 & 10 \\
\hline 2 . & \begin{array}{l}
-7 n^2+2 n
\end{array} & \begin{array}{l}
\begin{aligned}
&-7(2 n-1)+2 \\
&=-14 n+9
\end{aligned}
\end{array} & \begin{array}{l}
-14
\end{array} \\
\hline 3 . & -9 n^2-4 n & \begin{array}{l}
\begin{aligned}
&-9(2 n-1)-4 \\
&=-18 n+5
\end{aligned}
\end{array} & -18 \\
\hline 4 . & 4 n^2-n & 4(2 n-1)-1=8 n-5 & 8 \\
\hline
\end{array}
\)
Ratio of Sums is Given
Case-I: If ratio of the sums of \(m\) and \(n\) terms of an AP is given by
\(
\frac{S_m}{S_n}=\frac{A m^2+B m}{A n^2+B n}
\)
where \(A, B\) are constants and \(A \neq 0\).
\(
\begin{aligned}
& \therefore \quad S_m=\left(A m^2+B m\right) k \text {, } \\
& S_n=\left(A n^2+B n\right) k \\
& \Rightarrow \quad t_m=S_m-S_{m-1}=[A(2 m-1)+B] k \\
& t_n=S_n-S_{n-1}=[A(2 n-1)+B] k \\
& \therefore \quad \frac{t_m}{t_n}=\frac{A(2 m-1)+B}{A(2 n-1)+B} \\
&
\end{aligned}
\)
Example 16: The ratio of sums of \(m\) and \(n\) terms of an AP is \(m^2: n^2\). The ratio of the \(m\) th and \(n\)th terms is
(a) \((2 m+1):(2 n-1)\)
(b) \(m: n\)
(c) \((2 m-1):(2 n-1)\)
(d) None of these
Solution:
\(
\text { (c) Here, } \quad \frac{S_m}{S_n}=\frac{m^2}{n^2} \quad[\because A=1, B=0]
\)
\(
\therefore \quad \frac{t_m}{t_n}=\frac{(2 m-1)}{(2 n-1)}
\)
\(
\Rightarrow \quad t_m: t_n=(2 m-1):(2 n-1)
\)
Case-II: If ratio of the sums of \(n\) terms of two AP’s is given by
\(
\frac{S_n}{S_n^{\prime}}=\frac{A n+B}{C n+D}
\)
where, \(A, B, C, D\) are constants and \(A, C \neq 0\)
\(
\therefore \quad S_n=n(A n+B) k, S_n^{\prime}=n(C n+D) k
\)
\(
\Rightarrow \quad t_n=[A(2 n-1)+B] k, t_n^{\prime}=[C(2 n-1)+D] k
\)
\(
\Rightarrow \quad d=t_n-t_{n-1}=2 A, d^{\prime}=t_n^{\prime}-t_{n-1}^{\prime}=2 C
\)
\(
\therefore \quad \frac{t_n}{t_n^{\prime}}=\frac{A(2 n-1)+B}{C(2 n-1)+D} \text { and } \frac{d}{d^{\prime}}=\frac{A}{C}
\)
Note If \(A=0, C=0\)
Then, \(\frac{S_n}{S_n^{\prime}}=\frac{B}{D} \Rightarrow \frac{t_n}{t_n^{\prime}}=\frac{B}{D}\) and \(\frac{d}{d^{\prime}}=\frac{0}{0}=\) not defined
Remark
If \(\frac{t_n}{t_n^{\prime}}=\frac{a n+b}{c n+d}\)
where, \(a, b, c, d\) are constants and \(a, c \neq 0\), then
\(
\frac{S_n}{S_n^{\prime}}=\frac{a\left(\frac{n+1}{2}\right)+b}{c\left(\frac{n+1}{2}\right)+d}
\)
Example 17: The sums of \(n\) terms of two arithmetic progressions are in the ratio \((7 n+1):(4 n+17)\). Find the ratio of their \(n\)th terms and also common differences.
Solution: Given,
Here, \(S_n: S_n^{\prime}=(7 n+1):(4 n+17)\)
Here, \(\quad A=7, B=1, C=4\) and \(D=17\)
\(\therefore \quad \frac{t_n}{t_n^{\prime}}=\frac{7(2 n-1)+1}{4(2 n-1)+17}=\frac{14 n-6}{8 n+13}\)
and \(\quad \frac{d}{d^{\prime}}=\frac{A}{C}=\frac{7}{4}\)
Hence, \(\quad t_n: t_n^{\prime}=(14 n-6):(8 n+13)\) and \(d: d^{\prime}=7: 4\)
Example 18: The sums of \(n\) terms of two AP’s are in the ratio \((3 n-13):(5 n+21)\). Find the ratio of their 24th terms.
Solution: Given, \(S_n: S_n^{\prime}=(3 n-13):(5 n+21)\)
Here, \(\quad A=3, B=-13, C=5\) and \(D=21\)
\(
\therefore \quad \frac{t_{24}}{t_{24}^{\prime}}=\frac{3(2 \times 24-1)-13}{5(2 \times 24-1)+21}=\frac{128}{256}=\frac{1}{2}
\)
\(
\therefore \quad t_{24}: t_{24}^{\prime}=1: 2
\)
Example 19: How many terms of the series \(20+19 \frac{1}{3}+18 \frac{2}{3}+\ldots\) must be taken to make 300 ? Explain the double answer.
Solution: Here, given series is an AP with first term \(a=20\) and the common difference, \(d=-\frac{2}{3}\).
Let the sum of \(n\) terms of the series be 300 .
Then,
\(
\begin{array}{lc}
\Rightarrow & 300=\frac{n}{2}\left\{2 \times 20+(n-1)\left(-\frac{2}{3}\right)\right\} \\
\Rightarrow & 300=\frac{n}{3}\{60-n+1\} \\
\Rightarrow & n^2-61 n+900=0 \\
\Rightarrow & (n-25)(n-36)=0 \\
\Rightarrow & n=25 \text { or } n=36
\end{array}
\)
\(\therefore\) Sum of 25 terms \(=\) Sum of 36 terms \(=300\)
Explanation of double answer
Here, the common difference is negative, therefore terms go on diminishing and \(t_{31}=20+(31-1)\left(\frac{-2}{3}\right)=0\) i.e., 31st term becomes zero. All terms after 31st term are negative. These negative terms ( \(\left.t_{32}, t_{33}, t_{34}, t_{35}, t_{36}\right)\) when added to positive terms \(\left(t_{26}, t_{27}, t_{28}, t_{29}, t_{30}\right)\), they cancel out each other i.e., sum of terms from 26 th to 36 th terms is zero. Hence, the sum of 25 terms as well as that of 36 terms is 300.
Example 20: Find the arithmetic progression consisting of 10 terms, if the sum of the terms occupying the even places is equal to 15 and the sum of those occupying the odd places is equal to \(12 \frac{1}{2}\).
Solution: Let the successive terms of an AP be \(t_1, t_2, t_3, \ldots, t_9, t_{10}\). By hypothesis,
\(
\begin{aligned}
& t_2+t_4+t_6+t_8+t_{10} =15 \\
\Rightarrow & \frac{5}{2}\left(t_2+t_{10}\right) & =15 \\
\Rightarrow & t_2+t_{10} =6 \\
\Rightarrow & (a+d)+(a+9 d) =6 \\
\Rightarrow & 2 a+10 d =6 \dots(i)
\end{aligned}
\)
\(
\begin{aligned}
\text { and } & t_1+t_3+t_5+t_7+t_9 =12 \frac{1}{2} \\
\Rightarrow & \frac{5}{2}\left(t_1+t_9\right) & =\frac{25}{2} \\
\Rightarrow & t_1+t_9 =5 \\
\Rightarrow & a+a+8 d =5 \\
\Rightarrow & 2 a+8 d =5 \dots(ii)
\end{aligned}
\)
From Eqs. (i) and (ii), we get \(d=\frac{1}{2}\) and \(a=\frac{1}{2}\) Hence, the AP is \(\frac{1}{2}, 1,1 \frac{1}{2}, 2,2 \frac{1}{2}, \ldots\)
Example 21: If \(N\), the set of natural numbers is partitioned into groups \(S_1=\{1\}, S_2=\{2,3\}\), \(S_3=\{4,5,6\}, \ldots\), find the sum of the numbers in \(S_{50}\).
Solution: The number of terms in the groups are \(1,2,3, \ldots\)
\(\because\) The number of terms in the 50 th group \(=50\)
\(\therefore \quad\) The last term of 1st group \(=1\)
The last term of 2 nd group \(=3=1+2\)
The last term of 3 rd group \(=6=1+2+3\)
\(\quad \quad \quad \quad \quad \quad \quad \quad \vdots \quad \vdots \quad \vdots \quad \vdots\)
\(
\begin{aligned}
& \text { The last term of } 49 \text { th group }=1+2+3+\ldots+49 \\
& \therefore \text { First term of } 50 \text { th group }=1+(1+2+3+\ldots+49) \\
& \qquad=1+\frac{49}{2}(1+49)=1226 \\
& \begin{aligned}
\therefore \quad S_{50} & =\frac{50}{2}\{2 \times 1226+(50-1) \times 1\} \\
& =25 \times 2501=62525
\end{aligned}
\end{aligned}
\)
Example 22: Find the sum of first 24 terms of on AP \(t_1, t_2, t_3, \ldots\), if it is known that
\(
t_1+t_5+t_{10}+t_{15}+t_{20}+t_{24}=225 .
\)
Solution: We know that, in an AP the sums of the terms equidistant from the beginning and end is always same and is equal to the sum of first and last term.
Then, \(t_1+t_{24}=t_5+t_{20}=t_{10}+t_{15}\)
but given
\(
t_1+t_5+t_{10}+t_{15}+t_{20}+t_{24}=225
\)
\(
\begin{aligned}
\left(t_1+t_{24}\right)+\left(t_5+t_{20}\right)+\left(t_{10}+t_{15}\right) & =225 \\
3\left(t_1+t_{24}\right) & =225 \\
t_1+t_{24} & =75 \\
S_{24}=\frac{24}{2}\left(t_1+t_{24}\right)=12 \times 75 & =900
\end{aligned}
\)
Example 23: If \((1+3+5+\ldots+p)+(1+3+5+\ldots+q)\) \(=(1+3+5+\ldots+r)\), where each set of parentheses contains the sum of consecutive odd integers as shown, then find the smallest possible value of \(p+q+r\) (where, \(p>6\) ).
Solution: We know that, \(1+3+5+\ldots+(2 n-1)=n^2\)
Thus, the given equation can be written as
\(
\begin{aligned}
\left(\frac{p+1}{2}\right)^2+\left(\frac{q+1}{2}\right)^2 & =\left(\frac{r+1}{2}\right)^2 \\
\Rightarrow \quad(p+1)^2+(q+1)^2 & =(r+1)^2
\end{aligned}
\)
Therefore, \((p+1, q+1, r+1)\) form a Pythagorean triplet as
\(
p>6 \Rightarrow p+1>7
\)
The first Pythagorean triplet containing a number \(>7\) is \((6,8,10)\).
\(
\begin{array}{rr}
\Rightarrow & p+1=8 q+1=6, r+1=10 \\
\Rightarrow & p+q+r=21
\end{array}
\)
Properties of Arithmetic Progression
Remark
Remark
Example 24: If \(S_1, S_2, S_3, \ldots, S_p\) are the sums of \(n\) terms of \(p\) AP’s whose first terms are \(1,2,3, \ldots, p\) and common differences are \(1,2,3, \ldots,(2 p-1)\) respectively, show that \(\quad S_1+S_2+S_3+\ldots .+S_p=\frac{1}{2} n p(n p+1)\).
Solution: Here, \(S_1=1+2+3+\ldots\) upto \(n\) terms \(=\frac{n(n+1)}{2}\)
\(
\begin{aligned}
S_2 & =2+5+8+\ldots \text { upto } n \text { terms }=\frac{n}{2}[2 \cdot 2+(n-1) 3] \\
& =\frac{n(3 n+1)}{2}
\end{aligned}
\)
Similarly, \(S_3=3+8+13+\ldots\) upto \(n\) terms \(=\frac{n(5 n+1)}{2}\), etc.
Now, \(S_1+S_2+S_3+\ldots+S_p\)
\(=\frac{n(n+1)}{2}+\frac{n(3 n+1)}{2}+\frac{n(5 n+1)}{2}+\ldots\) upto \(p\) terms
\(=\frac{n}{2}[(n+3 n+5 n+\ldots\) upto \(p\) terms \()\)
\(+(1+1+1+\ldots\) upto \(p\) terms \()]\)
\(=\frac{n}{2}\left[\frac{p}{2}(2 n+(p-1) 2 n)+p\right]\)
\(=\frac{n p}{2}[n+n(p-1)+1]=\frac{1}{2} n p(n p+1)\)
Example 25: Let \(\alpha\) and \(\beta\) be roots of the equation \(x^2-2 x+A=0\) and let \(\gamma\) and \(\delta\) be the roots of the equation \(x^2-18 x+B=0\). If \(\alpha<\beta<\gamma<\delta\) are in arithmetic progression, then find the values of \(A\) and \(B\).
Solution: \(\because \alpha, \beta, \gamma, \delta\) are in AP.
Let
\(
\beta=\alpha+d, \gamma=\alpha+2 d, \delta=\alpha+3 d, d>0
\)
[here, sum of \(\alpha, \beta, \gamma, \delta\) is not given]
\(
\begin{array}{lr}
\text { Given, } & \alpha+\beta=2, \alpha \beta=A \\
\Rightarrow & 2 \alpha+d=2, \alpha \beta=A \dots(i) \\
\text { and } & \gamma+\delta=18, \gamma \delta=B \\
\Rightarrow & 2 \alpha+5 d=18, \gamma \delta=B \dots(ii)
\end{array}
\)
From Eqs. (i) and (ii), we get
\(
\begin{aligned}
& d=4, \alpha=-1 \\
& \therefore \quad \beta=3, \gamma=7, \delta=11 \\
&
\end{aligned}
\)
\(
\begin{array}{ll}
\Rightarrow & A=\alpha \beta=(-1)(3)=-3 \\
\text { and } & B=\gamma \delta=(7)(11)=77
\end{array}
\)
Example 26: The digits of a positive integer having three digits are in AP 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 digit in the unit’s place be \(a-d\), digit in the ten’s place be \(a\) and the digit in the hundred’s place be \(a+d\).
\(
\begin{aligned}
& \text { Sum of digits }=a-d+a+a+d=15 \text { [Given] } \\
& \Rightarrow \quad 3 a=15 \\
& \therefore \quad a=5 \dots(i)
\end{aligned}
\)
\(
\begin{aligned}
\therefore \text { Original number } & =(a-d)+10 a+100(a+d) \\
& =111 a+99 d=555+99 d
\end{aligned}
\)
and number formed by reversing the digits
\(
\begin{aligned}
& =(a+d)+10 a+100(a-d) \\
& =111 a-99 d=555-99 d
\end{aligned}
\)
\(
\begin{aligned}
& \text { Given, }(555+99 d)-(555-99 d)=594 \Rightarrow 198 d=594 \\
& \therefore \quad d=3
\end{aligned}
\)
Hence, original number \(=555+99 \times 3=852\)
Example 27: If three positive real numbers are in AP such that \(a b c=4\), then find the minimum value of \(b\).
Solution: \(\because a, b, c\) are in \(AP\).
Let \(a=A-D, b=A, c=A+D\)
Then, \(a=b-D, c=b+D\)
Now, \(a b c=4\)
\(
\Rightarrow \quad \begin{aligned}
(b-D) b(b+D) & =4 \\
b\left(b^2-D^2\right) & =4
\end{aligned}
\)
\(
\begin{array}{lr}
\Rightarrow & b^2-D^2<b^2 \\
\Rightarrow & b\left(b^2-D^2\right)<b^3 \Rightarrow 4<b^3
\end{array}
\)
\(
\therefore \quad b>(4)^{1 / 3} \text { or } b>(2)^{2 / 3}
\)
Hence, the minimum value of \(b\) is \((2)^{2 / 3}\).
Example 28: If \(a, b, c, d\) are distinct integers form an increasing AP such that \(d=a^2+b^2+c^2\), then find the value of \(a+b+c+d\).
Solution: Here, sum of numbers ie., \(a+b+c+d\) is not given.
Let \(\quad b=a+D, c=a+2 D, d=a+3 D, \forall D \in N\)
According to hypothesis,
\(
\begin{aligned}
& a+3 D=a^2+(a+D)^2+(a+2 D)^2 \\
\Rightarrow \quad & 5 D^2+3(2 a-1) D+3 a^2-a=0 \dots(i)
\end{aligned}
\)
\(
\begin{aligned}
\therefore \quad & =\frac{-3(2 a-1) \pm \sqrt{9(2 a-1)^2-20\left(3 a^2-a\right)}}{10} \\
& =\frac{-3(2 a-1) \pm \sqrt{\left(-24 a^2-16 a+9\right)}}{10}
\end{aligned}
\)
\(
\begin{array}{ll}
\text { Now, } & -24 a^2-16 a+9 \geq 0 \\
\Rightarrow & 24 a^2+16 a-9 \leq 0 \\
\Rightarrow & -\frac{1}{3}-\frac{\sqrt{70}}{3} \leq a \leq-\frac{1}{3}+\frac{\sqrt{70}}{12}
\end{array}
\)
\(
\Rightarrow \quad a=-1,0 \quad[\because a \in I]
\)
When \(a=0\) from Eq. (i), \(D=0, \frac{3}{5}\) (not possible \(\because D \in N\) ) and for \(a=-1\)
From Eq. (i),
\(
D=1, \frac{4}{5}
\)
\(
\therefore \quad D=1 \quad[\because D \in N]
\)
\(
\therefore \quad a=-1, b=0, c=1, d=2
\)
Then, \(\quad a+b+c+d=-1+0+1+2=2\)
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