0 of 74 Questions completed
Questions:
You have already completed the quiz before. Hence you can not start it again.
Quiz is loading…
You must sign in or sign up to start the quiz.
You must first complete the following:
0 of 74 Questions answered correctly
Your time:
Time has elapsed
You have reached 0 of 0 point(s), (0)
Earned Point(s): 0 of 0, (0)
0 Essay(s) Pending (Possible Point(s): 0)
For any three positive real numbers \(a , b\) and \(c\), \(9\left(25 a^2+b^2\right)+25\left(c^2-3 a c\right)=15 b(3 a+c)\). Then [JEE Main 2017]
We have
\(
\begin{aligned}
& 9\left(25 a ^2+ b ^2\right)+25\left( c ^2-3 ac \right)=15 b (3 a + c ) \\
& \Rightarrow 225 a ^2+9 b ^2+25 c ^2-75 ac =45 ab +15 bc \\
& \Rightarrow(15 a )^2+(3 b )^2+(5 c )^2-75 ac -45 ab -15 bc =0 \\
& \frac{1}{2}\left[(15 a -3 b )^2+(3 b -5 c )^2+(5 c -15 a )^2\right]=0
\end{aligned}
\)
it is possible when \(15 a -3 b =0,3 b -5 c =0\) and
\(
\begin{aligned}
& 5 c-15 a=0 \\
& \Rightarrow 15 a=3 b=5 \\
& \Rightarrow b=\frac{5 c}{3}, a=\frac{c}{3} \\
& \Rightarrow a+b=\frac{c}{3}+\frac{5 c}{3}=\frac{6 c}{3} \\
& \Rightarrow a+b=2 c \\
& \Rightarrow b, c, a \text { are in A.P. }
\end{aligned}
\)
If three positive numbers \(a, b\) and \(c\) are in A.P. such that abc \(=8\), then the minimum possible value of \(b\) is: [Online April 9, 2017]
\(
\begin{aligned}
& \text { A.M. } \geqslant \text { G. M. } \\
& \text { A.M. of } a, b, c=b \\
& \text { G.M. of } a, b, c=\sqrt[3]{a b c} \\
& \quad b \geqslant \sqrt[3]{a b c} \\
& \quad \Rightarrow b \geqslant \sqrt[3]{8} \quad(\because a b c=8) \\
& \Rightarrow \quad b \geqslant 2 \\
& \therefore \text { Minimum value of } b=2
\end{aligned}
\)
Let \(a_1, a_2, a_3, \ldots ., a_n\), be in A.P. If \(a_3+a_7+a_{11}+a_{15}=72\), then the sum of its first 17 terms is equal to: [Online April 10, 2016]
\(
\begin{aligned}
& a_3+a_7+a_{11}+a_{15}=72 \\
& \left(a_3+a_{15}\right)+\left(a_7+a_{11}\right)=72 \\
& a_3+a_{15}+a_7+a_{11}=2\left(a_1+a_{17}\right) \\
& a_1+a_{17}=36 \\
& S_{17}=\frac{17}{2}\left[a_1+a_{17}\right]=17 \times 18=306
\end{aligned}
\)
Let \(\alpha\) and \(\beta\) be the roots of equation \(px ^2+ qx + r =0, p \neq 0\). If \(p , q , r\) are in A.P and \(\frac{1}{\alpha}+\frac{1}{\beta}=4\), then the value of \(|\alpha-\beta|\) is: [JEE 2014]
Let \(p, q, r\) are in \(AP\)
\(
\Rightarrow 2 q=p+r \dots(i)
\)
Given \(\frac{1}{\alpha}+\frac{1}{\beta}=4\)
\(
\Rightarrow \frac{\alpha+\beta}{\alpha \beta}=4
\)
We have \(\alpha+\beta=-q / p\) and \(\alpha \beta=\frac{r}{p}\)
\(
\Rightarrow \frac{-\frac{q}{p}}{\frac{r}{p}}=4 \Rightarrow q=-4 r \dots(ii)
\)
From (i), we have
\(
2(-4 r)=p+r
\)
\(
\begin{aligned}
&\begin{aligned}
& p=-9 r \\
& q=-4 r \\
& r=r
\end{aligned}\\
&\begin{aligned}
& \text { Now }|\alpha-\beta|=\sqrt{(\alpha+\beta)^2-4 \alpha \beta} \\
& =\sqrt{\left(\frac{-q}{p}\right)^2-\frac{4 r}{p}}=\frac{\sqrt{q^2-4 p r}}{|p|} \\
& =\frac{\sqrt{16 r^2+36 r^2}}{|-9 r|}=\frac{2 \sqrt{13}}{9}
\end{aligned}
\end{aligned}
\)
The sum of the first 20 terms common between the series 3 \(+7+11+15+\) \(\ldots\) and \(1+6+11+16+\) \(\ldots\) is [Online April 11, 2014]
\(
\begin{aligned}
& \text { Given } n =20 ; S _{20}=\text { ? } \\
& \text { Series(1) } \rightarrow 3,7, \underline{11}, 15,19,23,27,\underline{31},35,39,43,47,\underline{51},55,59 \ldots
\end{aligned}
\)
\(
\begin{array}{r}
\text { Series }(2) \rightarrow 1,6,\underline{11},16,21,26,\underline{31},36,41,46,\underline{51},56,61,66,71.
\end{array}
\)
The common terms between both the series are \(11,31,51,71 \ldots\)
Above series forms an Arithmetic progression (A.P).
Therefore, first term (a) \(=11\) and common difference \(( d )=20\)
Now, \(S _{ n }=\frac{n}{2}[2 a+(n-1) d]\)
\(
\begin{aligned}
& S _{20}=\frac{20}{2}[2 \times 11+(20-1) 20] \\
& S _{20}=10[22+19 \times 20] \\
& S _{20}=10 \times 402=4020 \\
& \therefore S _{20}=4020
\end{aligned}
\)
Given an A.P. whose terms are all positive integers. The sum of its first nine terms is greater than 200 and less than 220 . If the second term in it is 12 , then its \(4^{\text {th }}\) term is: [Online April 9, 2014]
Let \(a\) be the first term and \(d\) be the common difference of given A.P.
Second term, \(a+d=12 \dots(1)\)
Sum of first nine terms,
\(
S _9=\frac{9}{2}(2 a+8 d)=9(a+4 d)
\)
Given that \(S _9\) is more than 200 and less than 220
\(
\begin{aligned}
& \Rightarrow 200< S _9<220 \\
& \Rightarrow 200<9(a+4 d)<220 \\
& \Rightarrow 200<9(a+d+3 d)<220
\end{aligned}
\)
Putting value of \((a+d)\) from equation (1)
\(
\begin{aligned}
& 200<9(12+3 d)<220 \\
& \Rightarrow 200<108+27 d<220 \\
& \Rightarrow 200-108<108+27 d-108<220-108 \\
& \Rightarrow 92<27 d<112
\end{aligned}
\)
Possible value of \(d\) is 4
\(
27 \times 4=108
\)
Thus, \(92<108<112\)
Putting value of \(d\) in equation (1)
\(
\begin{aligned}
& a+d=12 \\
& a=12-4=8 \\
& 4^{\text {th }} \text { term }=a+3 d=8+3 \times 4=20
\end{aligned}
\)
If \(a_1, a_2, a_3, \ldots, a_n, \ldots\) are in A.P. such that \(a_4-a_7+a_{10}=m\), then the sum of first 13 terms of this A.P., is: [Online April 23, 2013]
If \(d\) be the common difference, then
\(
\begin{aligned}
m & =a_4-a_7+a_{10}=a_4-a_7+a_7+3 d =a_7 \\
S _{13} & =\frac{13}{2}\left[a_1+a_{13}\right]=\frac{13}{2}\left[a_1+a_7+6 d\right] \\
& =\frac{13}{2}\left[2 a_7\right]=13 a_7=13 m
\end{aligned}
\)
Given sum of the first \(n\) terms of an A.P. is \(2 n+3 n^2\). Another A.P. is formed with the same first term and double of the common difference, the sum of \(n\) terms of the new A.P. is : [Online April 22, 2013]
Given \(S _n=2 n+3 n^2\)
Now, first term \(=2+3=5\)
second term \(=2(2)+3(4)=16\)
third term \(=2(3)+3(9)=33\)
Now, sum given in option (b) only has the same first term and difference between 2 nd and 1st term is double also.
Let \(a_1, a_2, a_3, \ldots\) be an A.P, such that \(\frac{a_1+a_2+\ldots+a_p}{a_1+a_2+a_3+\ldots+a_q}=\frac{p^3}{q^3} ; p \neq q\). Then \(\frac{a_6}{a_{21}}\) is equal to: [Online April 9, 2013]
\(
\frac{a_1+a_2+a_3+\ldots . .+a_p}{a_1+a_2+a_3+\ldots . .+a_q}=\frac{p^3}{q^3}
\)
\(
\begin{aligned}
& \frac{a_1+a_2}{a_1}=\frac{8}{1} \Rightarrow a_1+\left(a_1+d\right)=8 a_1 \\
& d=6 a_1
\end{aligned}
\)
Now \(\frac{a_6}{a_{21}}=\frac{a_1+5 d}{a_1+20 d}\)
\(
=\frac{a_1+5 \times 6 a_1}{a_1+20 \times 6 a_1}=\frac{1+30}{1+120}=\frac{31}{121}
\)
If 100 times the \(100^{\text {th }}\) term of an \(AP\) with non zero common difference equals the 50 times its \(50^{\text {th }}\) term, then the \(150^{\text {th }}\) term of this AP is : [JEE 2012]
Let \(100^{\text {th }}\) term of an \(A P\) is \(a+(100-1) d\) \(=a+99 d\) where ‘ \(a\) ‘ is the first term of \(A . P\) and ‘ \(d\) ‘ is the common difference of A.P.
Similarly, \(50^{\text {th }}\) term \(=a+(50-1) d\)
\(
=a+49 d
\)
Now, According to the question
\(
\begin{aligned}
& 100(a+99 d)=50(a+49 d) \\
& \Rightarrow \quad 2 a+198 d=a+49 d \Rightarrow a+149 d=0
\end{aligned}
\)
This is the \(150^{\text {th }}\) term of an A.P.
Hence, \(T_{150}=a+149 d=0\)
If the A.M. between \(p^{\text {th }}\) and \(q^{\text {th }}\) terms of an A.P. is equal to the A.M. between \(r^{\text {th }}\) and \(s^{\text {th }}\) terms of the same A.P., then \(p+q\) is equal to [Online May 26, 2012]
\(
\begin{aligned}
& \text { Given: } \frac{a_p+a_q}{2}=\frac{a_r+a_s}{2} \\
& \Rightarrow a+(p-1) d+a+(q-1) d \\
& \quad=a+(r-1) d+a+(s-1) d \\
& \Rightarrow 2 a+(p+q) d-2 d=2 a+(r+s) d-2 d \\
& \Rightarrow(p+q) d=(r+s) d \Rightarrow p+q=r+s .
\end{aligned}
\)
Suppose \(\theta\) and \(\phi(\neq 0)\) are such that \(\sec (\theta+\phi), \sec \theta\) and \(\sec (\theta-\phi)\) are in A.P. If \(\cos \theta=k \cos \left(\frac{\phi}{2}\right)\) for some \(k\), then \(k\) is equal to [Online May 19, 2012]
\(
\begin{aligned}
& \text { Since, } \sec (\theta-\phi), \sec \theta \text { and } \sec (\theta+\phi) \text { are in A.P., } \\
& \therefore 2 \sec \theta=\sec (\theta-\phi)+\sec (\theta+\phi) \\
& \Rightarrow \quad \frac{2}{\cos \theta}=\frac{\cos (\theta+\phi)+\cos (\theta-\phi)}{\cos (\theta-\phi) \cos (\theta+\phi)} \\
& \Rightarrow 2\left(\cos ^2 \theta-\sin ^2 \phi\right)=\cos \theta[2 \cos \theta \cos \phi] \\
& \Rightarrow \cos ^2 \theta(1-\cos \phi)=\sin ^2 \phi=1-\cos ^2 \phi \\
& \Rightarrow \quad \cos ^2 \theta=1+\cos \phi=2 \cos ^2 \frac{\phi}{2} \\
& \therefore \quad \cos \theta=\sqrt{2} \cos \frac{\phi}{2} \\
& \text { But given } \cos \theta=k \cos \frac{\phi}{2} \\
& \therefore \quad k=\sqrt{2}
\end{aligned}
\)
Let \(a_n\) be the \(n^{\text {th }}\) term of an A.P. If \(\sum_{r=1}^{100} a_{2 r}=\alpha\) and \(\sum_{r=1}^{100} a_{2 r-1}=\beta\), then the common difference of the A.P. is [JEE 2011]
\(
\begin{aligned}
& \text { Let A.P. be } a, a+d, a+2 d, \ldots . . . \\
& \qquad a_2+a_4+\ldots \ldots \ldots+a_{200}=\alpha
\end{aligned}
\)
\(
\frac{100}{2}[2(a+d)+(100-1) d]=\alpha \ldots \text { (i) }
\)
and \(a_1+a_3+a_5+\ldots \ldots \ldots .+a_{199}=\beta\)
\(
\frac{100}{2}[2 a+(100-1) d]=\beta \ldots \text { (ii) }
\)
On solving (i) and (ii), we get
\(
d=\frac{\alpha-\beta}{100}
\)
A man saves \(₹ 200\) in each of the first three months of his service. In each of the subsequent months his saving increases by ₹ 40 more than the saving of immediately previous month. His total saving from the start of service will be \(₹ 11040\) after [JEE 2011]
\(
\begin{aligned}
& \text { Let required number of months }= n \\
& \therefore 200 \times 3+\left(240+280+320+\ldots+( n -3)^{\text {th }} \text { term }\right) \\
& =11040 \\
& \quad \Rightarrow \frac{n-3}{2}[2 \times 240+(n-4) \times 40]=11040-600 \\
& \Rightarrow(n-3)[240+20 n-80]=10440 \\
& \Rightarrow(n-3)(20 n+160)=10440 \\
& \Rightarrow(n-3)(n+8)=522 \\
& \Rightarrow n^2+5 n-546=0 \\
& \Rightarrow(n+26)(n-21)=0 \\
& \therefore n=21
\end{aligned}
\)
A person is to count 4500 currency notes. Let \(a_n\) denote the number of notes he counts in the \(n ^{\text {th }}\) minute. If \(a_1=a_2=\ldots=\) \(a_{10}=150\) and \(a_{10}, a_{11}, \ldots\) are in an AP with common difference -2 , then the time taken by him to count all notes is [JEE 2010]
Till \(10^{\text {th }}\) minute number of counted notes \(=1500\)
\(
\begin{aligned}
3000 & =\frac{n}{2}[2 \times 148+(n-1)(-2)]=n[148-n+1] \\
& n^2-149 n+3000=0 \\
& \Rightarrow n=125,24
\end{aligned}
\)
But \(n=125\) is not possible
\(
\therefore \text { total time }=24+10=34 \text { minutes. }
\)
Let \(a_1, a_2, a_3\) \(\ldots\) be terms on A.P. If \(\frac{a_1+a_2+\ldots \ldots a_p}{a_1+a_2+\ldots \ldots \ldots+a_q}=\frac{p^2}{q^2}, p \neq q\), then \(\frac{a_6}{a_{21}}\) equals [JEE 2006]
\(
\begin{aligned}
& \frac{\frac{p}{2}\left[2 a_1+(p-1) d\right]}{\frac{q}{2}\left[2 a_1+(q-1) d\right]}=\frac{p^2}{q^2} \\
& \Rightarrow \frac{2 a_1+(p-1) d}{2 a_1+(q-1) d}=\frac{p}{q} \\
& \frac{a_1+\left(\frac{p-1}{2}\right) d}{a_1+\left(\frac{q-1}{2}\right) d}=\frac{p}{q} \\
& \text { For } \frac{a_6}{a_{21}}, p=11, q=41 \Rightarrow \frac{a_6}{a_{21}}=\frac{11}{41}
\end{aligned}
\)
If the coefficients of \(r\) th, \((r+1)\) th, and \((r+2)\) th terms in the the binomial expansion of \((1+y)^m\) are in A.P., then \(m\) and \(r\) satisfy the equation [JEE 2005]
\(
\text { Given }{ }^m C_{r-1},{ }^m C_r,{ }^m C_{r+1} \text { are in A.P. }
\)
\(
\begin{aligned}
& 2^m C_r={ }^m C_{r-1}+{ }^m C_{r+1} \\
& \Rightarrow 2=\frac{{ }^m C_{r-1}}{{ }^m C_r}+\frac{{ }^m C_{r+1}}{{ }^m C_r}=\frac{r}{m-r+1}+\frac{m-r}{r+1} \\
& \Rightarrow \quad m^2-m(4 r+1)+4 r^2-2=0 .
\end{aligned}
\)
Let \(T_{ r }\) be the rth 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 [JEE 2004]
\(
\begin{aligned}
& T_m=a+(m-1) d=\frac{1}{n} \dots(1) \\
& T_n=a+(n-1) d=\frac{1}{m} \dots(2)
\end{aligned}
\)
(1)-(2) \(\Rightarrow(m-n) d=\frac{1}{n}-\frac{1}{m} \Rightarrow d=\frac{1}{m n}\)
From(1) \(a=\frac{1}{m n} \Rightarrow a-d=0\)
If \(1, \log _9\left(3^{1-x}+2\right), \log _3\left(4.3^x-1\right)\) are in A.P. then \(x\) equals [JEE 2002]
\(
1, \log _9\left(3^{1-x}+2\right), \log _3\left(4.3^x-1\right) \text { are in A.P. }
\)
\(
\begin{aligned}
& 2 \log _9\left(3^{1-x}+2\right)=1+\log _3\left(4.3^x-1\right) \\
& \log _3\left(3^{1-x}+2\right)=\log _3 3+\log _3\left(4.3^x-1\right) \\
& \log _3\left(3^{1-x}+2\right)=\log _3\left[3\left(4.3^x-1\right)\right] \\
& 3^{1-x}+2=3\left(4.3^x-1\right) \\
& 3.3^{-x}+2=12.3^x-3 .
\end{aligned}
\)
Put \(3^x=t\)
\(\Rightarrow \frac{3}{t}+2=12 t-3\) or \(12 t^2-5 t-3=0\);
Hence \(t=-\frac{1}{3}, \frac{3}{4}\)
\(\Rightarrow 3^x=\frac{3}{4}\) (as \(\left.3^x \neq-v e\right)\)
\(\Rightarrow x=\log _3\left(\frac{3}{4}\right)\) or \(x=\log _3 3-\log _3 4\)
\(\Rightarrow x=1-\log _3 4\)
If the \(2^{\text {nd }}, 5^{\text {th }}\) and \(9^{\text {th }}\) terms of a non-constant A.P. are in G.P., then the common ratio of this G.P. is: (JEE 2016)
Let the GP be \(a\), ar and \(ar ^2\) then \(a = A + d ; ar = A +4 d\); \(ar ^2= A +8 d\)
\(\Rightarrow \frac{ ar ^2- ar }{ ar – a }=\frac{( A +8 d )-( A +4 d )}{( A +4 d )-( A + d )}\)
\(
r =\frac{4}{3}
\)
Let \(z=1+\) ai be a complex number, \(a>0\), such that \(z^3\) is areal number. Then the sum \(1+z+z^2+\ldots .+z^{11}\) is equal to: (Online April 10, 2016)
\(
\begin{aligned}
& z=1+a i \\
& z^2=1-a^2+2 a i \\
& z^2 \cdot z=\left\{\left(1-a^2\right)+2 a i\right\} \quad\{1+a i\} \\
& =\left(1-a^2\right)+2 a i+\left(1-a^2\right) \quad a i-2 a^2 \\
& z^3 \text { is real } \Rightarrow 2 a+\left(1-a^2\right) a=0 \\
& a\left(3-a^2\right)=0 \Rightarrow a=\sqrt{3} \quad(a>0) \\
& 1+z+z^2 \ldots \ldots . z^{11}=\frac{z^{12}-1}{z-1}=\frac{(1+\sqrt{3} i)^{12}-1}{1+\sqrt{3} i-1} \\
& =\frac{(1+\sqrt{3} i)^{12}-1}{\sqrt{3} i} \\
& (1+\sqrt{3} i)^{12}=2^{12}\left(\frac{1}{2}+\frac{\sqrt{3}}{2} i\right)^{12} \\
& =2^{12}\left(\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}\right)^{12}=2^{12}(\cos 4 \pi+i \sin 4 \pi)=2^{12} \\
& \frac{2^{12}-1}{\sqrt{3} i}=\frac{4095}{\sqrt{3} i}=-\frac{4095}{3} \sqrt{3} i=-1365 \sqrt{3} i
\end{aligned}
\)
If \(m\) is the A.M. of two distinct real numbers \(l\) and \(n (l, n >1)\) and \(G _1, G _2\) and \(G _3\) are three geometric means between \(l\) and \(n\), then \(G _1^4+2 G _2^4+ G _3^4\) equals. (JEE 2015)
Step 1: Solve for value of \(m\).
Given, \(m\) is the arithmetic mean of \(l\) and \(n\).
Then, \(m=\frac{l+n}{2}\)
\(
\Rightarrow 2 m=l+n
\)
Step 2: Solve for value of common ratio of G.P.
Now, \(l, G_1, G_2, G_3, n\) are in G.P.
Let \(r\) be the common ratio of this G.P.
\(
\therefore G_1=l r ; G_2=l r^2 ; G_3=l r^3 ; n=l r^4
\)
then, \(r^4=\frac{n}{l}\)
Step 3: Solve for value of \(G_1{ }^4+2 G_2{ }^4+G_3{ }^4\)
\(
\begin{aligned}
G_1{ }^4+2 G_2{ }^4+G_3{ }^4 & =(l r)^4+2\left(l r^2\right)^4+\left(l r^3\right)^4 \\
G_1{ }^4+2 G_2{ }^4+G_3{ }^4 & =l^4 r^4+2 l^4 r^8+l^4 r^{12} \\
G_1{ }^4+2 G_2{ }^4+G_3{ }^4 & =l^4\left[r^4+2 r^8+r^{12}\right]
\end{aligned}
\)
Substituting the values of \(r^4\), we get
\(
\begin{aligned}
& G_1{ }^4+2 G_2{ }^4+G_3{ }^4=l^4\left[\frac{n}{l}+2 \frac{n^2}{l^2}+\frac{n^3}{l^3}\right] \\
& G_1{ }^4+2 G_2{ }^4+G_3{ }^4=n l^3\left[1+\frac{2 n}{l}+\frac{n^2}{l^2}\right] \\
& =n l^3\left[1+\frac{n}{l}\right]^2 \\
& =n l^3\left(\frac{(l+n)^2}{l^2}\right) \\
& =n l(l+n)^2 \\
& =n l(2 m)^2 \\
& =4 m^2 n l
\end{aligned}
\)
The sum of the \(3^{\text {rd }}\) and the \(4^{\text {th }}\) terms of a G.P. is 60 and the product of its first three terms is 1000 . If the first term of this G.P. is positive, then its \(7^{\text {th }}\) term is: [Online April 11, 2015]
Let \(a , ar\) and \(ar ^2\) be the first three terms of G.P According to the question
\(
a ( ar )\left( ar ^2\right)=1000 \Rightarrow( ar )^3=1000 \Rightarrow ar =10
\)
\(
\begin{aligned}
& \text { and } ar ^2+ ar ^3=60 \Rightarrow ar \left( r + r ^2\right)=60 \\
& \Rightarrow r ^2+ r -6=0 \\
& \Rightarrow r =2,-3
\end{aligned}
\)
\(a=5, a=-\frac{10}{3}\) (reject)
Hence, \(T _7= ar ^6=5(2)^6=5 \times 64=320\).
Three positive numbers form an increasing G. P. If the middle term in this GP. is doubled, the new numbers are in A.P. then the common ratio of the G.P. is: [JEE 2014]
Let \(a, a r, a r^2\) are in G.P.
According to the question \(a, 2 a r, a r^2\) are in A.P.
\(
\Rightarrow \quad 2 \times 2 a r=a+a r^2
\)
\(
\begin{gathered}
\Rightarrow \quad 4 r=1+r^2 \Rightarrow r^2-4 r+1=0 \\
r=\frac{4 \pm \sqrt{16-4}}{2}=2 \pm \sqrt{3}
\end{gathered}
\)
Since \(r>1\)
\(\therefore \quad r=2-\sqrt{3}\) is rejected
Hence, \(r=2+\sqrt{3}\)
The least positive integer \(n[latex] such that [latex]1-\frac{2}{3}-\frac{2}{3^2}-\ldots .-\frac{2}{3^{ n -1}}<\frac{1}{100}\), is: [Online April 12, 2014]
\(
\begin{aligned}
& 1-\frac{2}{3}-\frac{2}{3^2} \ldots . \frac{2}{3^{n-1}}<\frac{1}{100} \\
& \Rightarrow 1-\frac{2}{3}\left[\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\ldots \frac{1}{3^{n-1}}\right]<\frac{1}{100}
\end{aligned}
\)
\(
\begin{aligned}
& \frac{1-2\left[\frac{1}{3}\left(\frac{1}{3^n}-1\right)\right]}{\frac{1}{3}-1}<\frac{1}{100} \\
\Rightarrow & 1-2\left[\frac{3^n-1}{2.3^n}\right]<\frac{1}{100} \\
\Rightarrow & 1-\left[\frac{3^n-1}{3^n}\right]<\frac{1}{100}
\end{aligned}
\)
\(
\Rightarrow 100<3^n
\)
Thus, least value of \(n\) is 5.
In a geometric progression, if the ratio of the sum of first 5 terms to the sum of their reciprocals is 49 , and the sum of the first and the third term is 35 . Then the first term of this geometric progression is: [Online April 11, 2014]
Let the sum of first 5 terms of the GP be
\(
a / r^2+a / r+a+a r+a r^2
\)
Now according to question
\(
\begin{aligned}
& \frac{\frac{a}{r^2}+\frac{a}{r}+a+a r+a r^2}{\frac{r^2}{a}+\frac{r}{a}+\frac{1}{a}+\frac{1}{a r}+\frac{1}{a r^2}} \\
& \quad=49 \Rightarrow a^2=49 \Rightarrow a= \pm 7
\end{aligned}
\)
\(
\begin{aligned}
& \frac{7}{r^2}+7=35 \Rightarrow \frac{7}{r^2}=28 \\
& \Rightarrow r^2=\frac{1}{4} \Rightarrow r= \pm \frac{1}{2} \\
&
\end{aligned}
\)
Thus, first term is
\(
\frac{7}{\left(\frac{1}{2}\right)^2}=7 \times 4=28
\)
The coefficient of \(x^{50}\) in the binomial expansion of
\(
\begin{aligned}
& (1+x)^{1000}+x(1+x)^{999}+x^2(1+x)^{998}+\ldots \\
& +x^{1000} \text { is: } \\
& \text { [Online April 11, 2014] }
\end{aligned}
\) [Online April 11, 2014]
Let given expansion be
\(
\begin{aligned}
S =(1+x)^{1000}+x(1+x)^{999}+x^2(1 & +x)^{998}+\ldots +\ldots+x^{1000}
\end{aligned}
\)
Put \(1+x=t\)
\(
S =t^{1000}+x t^{999}+x^2(t)^{998}+\ldots+x^{1000}
\)
This is a G.P with common ratio \(\frac{x}{t}\)
\(
\begin{aligned}
& S =\frac{t^{1000}\left[1-\left(\frac{x}{t}\right)^{1001}\right]}{1-\frac{x}{t}} \\
& =\frac{(1+x)^{1000}\left[1-\left(\frac{x}{1+x}\right)^{1001}\right]}{1-\frac{x}{1+x}} \\
& =\frac{(1+x)^{1001}\left[(1+x)^{1001}-x^{1001}\right]}{(1+x)^{1001}} \\
& =\left[(1+x)^{1001}-x^{1001}\right]
\end{aligned}
\)
Now coeff of \(x^{50}\) in above expansion is equal to coeff of \(x^{50}\) in \((1+x)^{1001}\) which is \({ }^{1001} C_{50}\)
\(
=\frac{(1001)!}{50!(951)!}
\)
Given a sequence of 4 numbers, first three of which are in G.P. and the last three are in A.P. with common difference six. If first and last terms of this sequence are equal, then the last term is: [Online April 25, 2013]
Let \(a, b, c, d\) be four numbers of the sequence.
Now, according to the question \(b^2=a c\) and \(c-b=6\) and \(a-c=6\)
Also, given \(a=d\)
\(
\begin{aligned}
& \therefore b^2=a c \Rightarrow b^2=a\left[\frac{a+b}{2}\right] \quad(\because 2 c=a+b) \\
& \Rightarrow a^2-2 b^2+a b=0
\end{aligned}
\)
Now, \(c-b=6\) and \(a-c=6\),
gives \(a-b=12\)
\(
\begin{aligned}
& \Rightarrow b=a-12 \\
& \therefore a^2-2 b^2+a b=0 \\
& \Rightarrow a^2-2(a-12)^2+a(a-12)=0 \\
& \Rightarrow a^2-2 a^2-288+48 a+a^2-12 a=0 \\
& \Rightarrow 36 a=288 \Rightarrow a=8
\end{aligned}
\)
Hence, last term is \(d=a=8\).
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 [Online May 12, 2012]
The given relation can be written as
\(
\begin{aligned}
& \left(a^2 p^2-2 a b p+b^2\right)+\left(b^2 p^2+c^2-2 b p c\right)+ \\
& \left(c^2 p^2+d^2-2 p c d\right) \leq 0 \\
& \text { or } \quad(a p-b)^2+(b p-c)^2+(c p-d)^2 \leq 0 \dots(1)
\end{aligned}
\)
Since \(a, b, c , d\) and \(p\) are all real, the inequality (1) is possible only when each of factor is zero.
i.e., \(a p-b=0, b p-c=0\) and \(c p-d=0\)
or \(\quad p=\frac{b}{a}=\frac{c}{b}=\frac{d}{c}\)
or \(a, b, c, d\) are in G.P.
The difference between the fourth term and the first term of a Geometrical Progression is 52 . If the sum of its first three terms is 26 , then the sum of the first six terms of the progression is [Online May 7, 2012]
Let \(a, a r, a r^2, a r^3, a r^4, a r^5\) be six terms of a G.P. where ‘ \(a\) ‘ is first term and \(r\) is common ratio.
According to given conditions, we have
\(
a r^3-a=5 \Rightarrow a\left(r^3-1\right)=52 \dots(1)
\)
\(
\begin{aligned}
& \text { and } a+a r+a r^2=26 \\
& \Rightarrow a\left(1+r+r^2\right)=26 \dots(2)
\end{aligned}
\)
To find: \(a\left(1+r+r^2+r^3+r^4+r^5\right)\) Consider
\(
\begin{aligned}
& a\left[1+r+r^2+r^3+r^4+r^5\right] \\
& =a\left[1+r+r^2+r^3\left(1+r+r^2\right)\right] \\
& =a\left[1+r+r^2\right]\left[1+r^3\right] \dots(3)
\end{aligned}
\)
Divide (1) by (2), we get
\(
\frac{r^3-1}{1+r+r^2}=2 \text {, }
\)
we know \(r^3-1=(r-1)\left(1+r+r^2\right)\)
\(
\begin{aligned}
& \therefore r-1=2 \Rightarrow r=3 \text { and } a=2 \\
& \therefore \quad a\left(1+r+r^2+r^3+r^4+r^5\right) \\
& =a\left(1+r+r^2\right)\left(1+r^3\right) \\
& =2(1+3+9)(1+27) \\
& =26 \times 28=728 \\
&
\end{aligned}
\)
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 alternately positive and negative, then the first term is [JEE 2008]
As per question,
\(
\begin{aligned}
& a+a r=12 \\
& a r^2+a r^3=48
\end{aligned}
\)
\(
\Rightarrow \frac{a r^2(1+r)}{a(1+r)}=\frac{48}{12} \Rightarrow r ^2=4, \Rightarrow r =-2 (\because \text { terms are }=+ \text { ve and }- \text { ve alternately })
\)
\(
\Rightarrow \quad a =-12
\)
In a geometric progression consisting of positive terms, each term equals the sum of the next two terms. Then the common ratio of its progression is equals [JEE 2007]
Let the series \(a , ar , ar ^2, \ldots .\). are in geometric progression. given, \(a = ar + ar ^2\)
\(
\begin{aligned}
& \Rightarrow \quad 1=r+r^2 \Rightarrow r^2+r-1=0 \\
& \Rightarrow \quad r=\frac{-1 \pm \sqrt{1-4 \times-1}}{2} \\
& \Rightarrow r=\frac{-1 \pm \sqrt{5}}{2}
\end{aligned}
\)
\(\Rightarrow r =\frac{\sqrt{5}-1}{2}[\because\) terms of G.P. are positive \(\therefore r\) should be positive]
The value of \(\sum_{k=1}^{10}\left(\sin \frac{2 k \pi}{11}+i \cos \frac{2 k \pi}{11}\right)\) is [JEE 2006]
\(
\begin{aligned}
& \sum_{k=1}^{10}\left(\sin \frac{2 k \pi}{11}+i \cos \frac{2 k \pi}{11}\right) \\
& =i \sum_{k=1}^{10}\left(\cos \frac{2 k \pi}{11}-i \sin \frac{2 k \pi}{11}\right) \\
& =i \sum_{k=1}^{10} e^{-\frac{2 k \pi}{11} i}=i\left\{\sum_{k=0}^{10} e^{-\frac{2 k \pi}{11} i}-1\right\} \\
& =i\left[1+e^{-\frac{2 \pi}{11} i}+e^{-\frac{4 \pi}{11} i}+\ldots .11 \text { terms }\right]-i \\
& =i\left[\frac{1-\left(e^{-\frac{2 \pi}{11}}\right)^{11}}{1-e^{-\frac{2 \pi}{11} i}}\right]-i=i\left[\frac{1-e^{-2 \pi i}}{1-e^{-\frac{2 \pi}{11} i}}\right]-i \\
& =i \times 0-i \quad\left[\because e^{-2 \pi i}=1\right] \\
& =-i \\
&
\end{aligned}
\)
If the expansion in powers of \(x\) of the function \(\frac{1}{(1-a x)(1-b x)}\) is \(a_0+a_1 x+a_2 x^2+a_3 x^3 \ldots .\). then \(a_n\) is [JEE 2006]
\(
\begin{aligned}
& (1-a x)^{-1}(1-b x)^{-1} \\
& =\left(1+a x+a^2 x^2+\ldots\right)\left(1+b x+b^2 x^2+\ldots\right)
\end{aligned}
\)
\(\therefore\) Coefficient of \(x ^{ n }\)
\(
x^n=b^n+a b^{n-1}+a^2 b^{n-2}+\ldots \ldots . .+a^{n-1} b+a^n
\)
{which is a G.P. with \(r=\frac{a}{b}\)
\(\therefore\) Its sum is \(\left.=\frac{b^n\left[1-\left(\frac{a}{b}\right)^{n+1}\right]}{1-\frac{a}{b}}\right\}\)
\(
=\frac{b^{n+1}-d^{n+1}}{b-a} \quad \therefore a_n=\frac{b^{n+1}-a^{n+1}}{b-a}
\)
Let two numbers have arithmetic mean 9 and geometric mean 4. Then these numbers are the roots of the quadratic equation [JEE 2004]
Let two numbers be \(a\) and \(b\) then \(\frac{a+b}{2}=9\) and \(\sqrt{a b}=4\)
\(\therefore\) Equation with roots a and \(b\) is
\(
x^2-(a+b) x+a b=0 \Rightarrow x^2-18 x+16=0
\)
Sum of infinite number of terms of GP is 20 and sum of their square is 100 . The common ratio of GP is [JEE 2002]
Let \(a=\) first term of G.P. and \(r=\) common ratio of G.P.; Then G.P. is \(a, a r, a r^2\)
Given \(S_{\infty}=20 \Rightarrow \frac{a}{1-r}=20\)
\(
\begin{aligned}
& \Rightarrow a=20(1-r) \ldots \text { (i) } \\
& \text { Also } a^2+a^2 r^2+a^2 r^4+\ldots \text { to } \infty=100
\end{aligned}
\)
\(
\begin{aligned}
& \Rightarrow \frac{a^2}{1-r^2}=100 \\
& \Rightarrow a^2=100(1-r)(1+r) \ldots(ii)
\end{aligned}
\)
From (i), \(a^2=400(1-r)^2\);
From (ii), we get \(100(1-r)(1+r)=400(1-r)^2\)
\(
\Rightarrow 1+r=4-4 r \Rightarrow 5 r=3 \Rightarrow r=3 / 5 \text {. }
\)
Fifth term of a GP is 2 , then the product of its 9 terms is [JEE 2002]
\(
\begin{aligned}
& a r^4=2 \\
& \quad a \times a r \times a r^2 \times a r^3 \times a r^4 \times a r^5 \times a r^6 \times a r^7 \times a r^8 \\
& =a^9 r^{36}=\left(a r^4\right)^9=2^9=512
\end{aligned}
\)
If the arithmetic mean of two numbers \(a\) and \(b, a>b>0\), is five times their geometric mean, then \(\frac{a+b}{a-b}\) is equal to : [Online April 8, 2017]
According to question
\(
\begin{aligned}
\text { A.M } & =5 \text { G.M. } \\
\frac{a+b}{2} & =5 \sqrt{a b} \\
\frac{a+b}{\sqrt{a b}} & =10 \\
\therefore \quad \frac{a}{b} & =\frac{10+\sqrt{96}}{10-\sqrt{96}}=\frac{10+4 \sqrt{6}}{10-4 \sqrt{6}}
\end{aligned}
\)
Use Componendo and Dividendo
\(
\frac{a+b}{a-b}=\frac{20}{8 \sqrt{6}}=\frac{5}{2 \sqrt{6}}=\frac{5 \sqrt{6}}{12}
\)
If \(A>0, B>0\) and \(A+B=\frac{\pi}{6}\), then the minimum value of \(\tan A +\tan B\) is : [Online April 10, 2016]
\(
\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B}
\)
\(
\begin{aligned}
& \frac{1}{\sqrt{3}}=\frac{y}{1-\tan A \tan B} \text { where } y=\tan A+\tan B \\
& \tan A \tan B=1-\sqrt{3} y \\
& \text { Also } AM \geq GM \\
& \frac{\tan A+\tan B}{2} \geq \sqrt{\tan A \tan B}
\end{aligned}
\)
\(
\begin{aligned}
& y \geq 2 \sqrt{1-\sqrt{3} y} \\
& y^2 \geq 4-4 \sqrt{3} y \\
& y^2+4 \sqrt{3} y-4 \geq 0 \\
& y \leq-2 \sqrt{3}-4 \text { or } y \geq-2 \sqrt{3}+4
\end{aligned}
\)
\(
(y \leq-2 \sqrt{3}-4 \text { is not possible as } \tan A \tan B>0)
\)
Let \(x , y , z\) be positive real numbers such that \(x + y + z =12\) and \(x^3 y^4 z^5=(0.1)(600)^3\). Then \(x^3+y^3+z^3\) is equal to: [Online April 9, 2016]
\(
\begin{aligned}
& x+y+z=12 \\
& AM \geq GM \\
& \frac{3\left(\frac{x}{3}\right)+4\left(\frac{y}{4}\right)+5\left(\frac{z}{5}\right)}{12} \geq \sqrt[12]{\left(\frac{x}{3}\right)^3\left(\frac{y}{4}\right)^4\left(\frac{z}{5}\right)^5} \\
& \frac{x^3 y^4 z^5}{3^3 4^4 5^5} \leq 1
\end{aligned}
\)
\(
\begin{aligned}
& x^3 y^4 z^5 \leq 3^3 .4^4 .5^5 \\
& x^3 y^4 z^5 \leq(0.1)(600)^3
\end{aligned}
\)
But, given \(x^3 y^4 z^5=(0.1)(600)^3\)
all the number are equal
\(
\begin{aligned}
& \frac{x}{3}=\frac{y}{4}=\frac{z}{5}(=k) \\
& x=3 k ; y=4 k ; z=5 k \\
& x+y+z=12 \\
& 3 k+4 k+5 k=12 \\
& k=1 \\
& x=3 ; y=4 ; z=5 \\
& x^3+y^3+z^3=216
\end{aligned}
\)
Let \(G\) be the geometric mean of two positive numbers \(a\) and \(b\), and \(M\) be the arithmetic mean of \(\frac{1}{ a }\) and \(\frac{1}{ b }\). If \(\frac{1}{ M }\) :G is \(4: 5\), then \(a : b\) can be: [Online April 12, 2014]
\(
\begin{aligned}
G & =\sqrt{a b} \\
M & =\frac{\frac{1}{a}+\frac{1}{b}}{2} \\
M & =\frac{a+b}{2 a b}
\end{aligned}
\)
Given that \(\frac{1}{ M }: G =4: 5\)
\(
\begin{aligned}
& \frac{2 a b}{(a+b) \sqrt{a b}}=\frac{4}{5} \\
& \Rightarrow \frac{a+b}{2 \sqrt{a b}}=\frac{5}{4} \\
& \Rightarrow \frac{a+b+2 \sqrt{a b}}{a+b-2 \sqrt{a b}}=\frac{5+4}{5-4} \text { \{Using Componendo \& Dividendo\} }
\end{aligned}
\)
\(
\begin{aligned}
& \sqrt{\frac{b}{a}}=\frac{4}{2}=2 \\
& \frac{b}{a}=\frac{4}{1} \\
& \frac{a}{b}=\frac{1}{4} \Rightarrow a: b=1: 4
\end{aligned}
\)
If \(a_1, a_2, \ldots \ldots . . ., a_n\) are in H.P., then the expression \(a_1 a_2+a_2 a_3+\ldots \ldots \ldots .+a_{n-1} a_n\) is equal to [JEE 2006]
\(
\frac{1}{a_2}-\frac{1}{a_1}=\frac{1}{a_3}-\frac{1}{a_2}=\ldots \ldots . . .=\frac{1}{a_n}-\frac{1}{a_{n-1}}=d \text { (say) }
\)
Then \(a_1 a_2=\frac{a_1-a_2}{d}, a_2 a_3=\frac{a_2-a_3}{d}\),
\(
\ldots \ldots \ldots, a_{n-1} a_n=\frac{a_{n-1}-a_n}{d}
\)
\(
\therefore \quad a_1 a_2+a_2 a_3+\ldots \ldots .+a_{n-1} a_n
\)
\(
\begin{aligned}
& =\frac{a_1-a_2}{d}+\frac{a_2-a_3}{d}+\ldots .+\frac{a_{n-1}-a_n}{d} \\
& =\frac{1}{d}\left[a_1-a_2+a_2-a_3+\ldots .+a_{n-1}-a_n\right]=\frac{a_1-a_n}{d}
\end{aligned}
\)
Also, \(\frac{1}{a_n}=\frac{1}{a_1}+(n-1) d\)
\(
\begin{aligned}
& \Rightarrow \frac{a_1-a_n}{a_1 a_n}=(n-1) d \\
& \Rightarrow \frac{a_1-a_n}{d}=(n-1) a_1 a_n
\end{aligned}
\)
Which is the required result.
If \(x=\sum_{n=0}^{\infty} a^n, y=\sum_{n=0}^{\infty} b^n, z=\sum_{n=0}^{\infty} c^n\) where \(a, b, c\) are in A.P
and \(| a |<1,|b|<1,|c|<1\) then \(x, y, z\) are in [JEE 2005]
\(
\begin{array}{ll}
x=\sum_{n=0}^{\infty} a^n=\frac{1}{1-a} & a=1-\frac{1}{x} \\
y=\sum_{n=0}^{\infty} b^n=\frac{1}{1-b} & b=1-\frac{1}{y} \\
z=\sum_{n=0}^{\infty} c^n=\frac{1}{1-c} & c=1-\frac{1}{z}
\end{array}
\)
\(a, b, c\) are in A.P. OR \(2 b=a+c\)
\(
2\left(1-\frac{1}{y}\right)=1-\frac{1}{x}+1-\frac{1}{y}
\)
\(\frac{2}{y}=\frac{1}{x}+\frac{1}{z} \Rightarrow x, y, z\) are in H.P.
If the sum of the roots of the quadratic equation \(a x^2+b x+c=0\) is equal to the sum of the squares of their reciprocals, then \(\frac{a}{c}, \frac{b}{a}\) and \(\frac{c}{b}\) are in [JEE 2003]
\(
a x^2+b x+c=0, \alpha+\beta=\frac{-b}{a}, \alpha \beta=\frac{c}{a}
\)
As for given condition, \(\alpha+\beta=\frac{1}{\alpha^2}+\frac{1}{\beta^2}\)
\(
\alpha+\beta=\frac{\alpha^2+\beta^2}{\alpha^2 \beta^2}-\frac{b}{a}=\frac{\frac{b^2}{a^2}-\frac{2 c}{a}}{\frac{c^2}{a^2}}
\)
On simplification \(2 a^2 c=a b^2+b c^2\)
\(\Rightarrow \frac{2 a}{b}=\frac{c}{a}+\frac{b}{c} \Rightarrow \frac{c}{a}, \frac{a}{b}, \frac{b}{c}\) are in A.P.
\(\therefore \frac{a}{c}, \frac{b}{a}, \& \frac{c}{b}\) are in H.P.
Let \(a, b, c \in R\) If \(f(x)=a x^2+b x+c\) is such that \(a+b+c=3\) and \(f(x+y)=f(x)+f(y)+x y, \forall x, y \in R\), then \(\sum_{n=1}^{10} f(n)\) is equal to: [JEE Main 2017]
\(
\begin{aligned}
& f(x)=a x^2+b x+c \\
& f(1)=a+b+c=3 \Rightarrow f(a)=3 \\
& \text { Now } f(x+y)=f(x)+f(y)+x y \\
& \text { Put } x=y=1 \text { in eqn (a) } \\
& f(2)=f(1)+f(1)+1 \\
& \quad=2 f(1)+1 \\
& f(2)=7 \\
& \Rightarrow f(3)=12
\end{aligned}
\)
\(
\text { Now, } S _{ n }=3+7+12+\ldots \ldots \ldots t _{ n } \dots(a)
\)
\(
S _{ n }=3+7+\ldots \ldots \ldots t _{ n -1}+ t _{ n } \dots(b)
\)
Subtract (b) from (a)
\(t _{ n }=3+4+5+\ldots[latex]. upto [latex]n\) terms
\(
\begin{aligned}
& t _{ n }=\frac{\left( n ^2+5 n \right)}{2} \\
& S _{ n }=\sum t _{ n }=\sum \frac{\left( n ^2+5 n \right)}{2} \\
& S _{ n }=\frac{1}{2}\left[\frac{ n ( n +1)(2 n +1)}{6}+\frac{5 n ( n +1)}{2}\right]=\frac{ n ( n +1)( n +8)}{6} \\
& S _{10}=\frac{10 \times 11 \times 18}{6}=330
\end{aligned}
\)
Let
\(
S_n=\frac{1}{1^3}+\frac{1+2}{1^3+2^3}+\frac{1+2+3}{1^3+2^3+3^3}+\ldots . .+\frac{1+2+\ldots \ldots .+n}{1^3+2^3+\ldots \ldots .+n^3}
\)
If \(100 S _n=n\), then \(n\) is equal to : [Online April 9, 2017]
\(
T_n=\frac{\frac{n(n+1)}{2}}{\left(\frac{n(n+1)}{2}\right)^2}
\)
\(
\begin{aligned}
& \Rightarrow \quad T_n=\frac{2}{n(n+1)} \\
& \Rightarrow \quad S_n=\sum T_n=2 \sum_{n=1}^n\left(\frac{1}{n}-\frac{1}{n+1}\right) \\
& =2\left\{1-\frac{1}{n+1}\right\} \\
& \Rightarrow \quad S_n=\frac{2 n}{n+1} \\
& \because \quad 100 S _{ n }= n \\
& \Rightarrow 100 \times \frac{2 n}{n+1}=n \\
& \Rightarrow n+1=200 \\
& \Rightarrow n=199 \\
&
\end{aligned}
\)
If the sum of the first \(n\) terms of the series \(\sqrt{3}+\sqrt{75}+\sqrt{243}\) \(+\sqrt{507}+\ldots \ldots\) is \(435 \sqrt{3}\), then \(n\) equals : [Online April 8, 2017]
Given,
\(
\begin{aligned}
& \sqrt{3}+\sqrt{75}+\sqrt{243}+\sqrt{507}+\ldots \ldots+n \text { terms } \\
& =\sqrt{3}+\sqrt{25 \times 3}+\sqrt{81 \times 3}+\sqrt{169 \times 3}+\ldots \ldots+n \text { terms } \\
& =\sqrt{3}+5 \sqrt{3}+9 \sqrt{3}+13 \sqrt{3}+\ldots \ldots+n \text { terms } \\
& =\sqrt{3}[1+5+9+13+\ldots \ldots+n \text { terms }] \\
& =\sqrt{3}\left[\frac{n}{2}(2.1+(n-1) 4)\right] \\
& =\sqrt{3}\left[\frac{n}{2}(2+4 n-4)\right] \\
& =\sqrt{3}\left[\frac{n}{2}(4 n-2)\right] \\
& =\sqrt{3}[n(2 n-1)]
\end{aligned}
\)
According to question,
\(
\begin{aligned}
& \sqrt{3}[n(2 n-1)]=435 \sqrt{3} \\
& \Rightarrow 2 n^2-n=435 \\
& \therefore n=\frac{1 \pm \sqrt{1+4 \times 2 \times 435}}{4}=\frac{1 \pm 59}{4} \\
& \therefore n=\frac{1+59}{4}=15 \text { or } \frac{1-59}{4}=-14.5 \\
& \therefore n=15 \text { (as } n \text { can’t be }-v e \text { ) }
\end{aligned}
\)
If the sum of the first ten terms of the series
\(
\left(1 \frac{3}{5}\right)^2+\left(2 \frac{2}{5}\right)^2+\left(3 \frac{1}{5}\right)^2+4^2+\left(4 \frac{4}{5}\right)^2+\ldots \ldots ., \text { is } \frac{16}{5} m \text {, }
\) then \(m\) is equal to: [JEE Main 2016]
\(
\begin{aligned}
& \left(\frac{8}{5}\right)^2+\left(\frac{12}{5}\right)^2+\left(\frac{16}{5}\right)^2+\left(\frac{20}{5}\right)^2 \ldots+\left(\frac{44}{5}\right)^2 \\
& S =\frac{16}{25}\left(2^2+3^2+4^2+\ldots+11^2\right) \\
& =\frac{16}{25}\left(\frac{11(11+1)(22+1)}{6}-1\right) \\
& =\frac{16}{25} \times 505=\frac{16}{5} \times 101 \\
& \Rightarrow \frac{16}{5} m =\frac{16}{5} \times 101 \\
& \Rightarrow m =101 .
\end{aligned}
\)
For \(x \in R, x \neq-1\), if \((1+x)^{2016}+x(1+x)^{2015}+x^2\) \((1+x)^{2014}+\ldots .+x^{2016}=\sum_{i=0}^{2016} a_i x^i\), then \(a_{17}\) is equal to : [Online April 9, 2016]
\(
S=(1+x)^{2016}+x(1+x)^{2015}+x^2(1+x)^{2014}+\ldots . .+x^{2015}(1+x)+x^{2016} \dots(i)
\)
\(
\left(\frac{x}{1+x}\right) S=x(1+x)^{2015}+x^2(1+x)^{2014}+\ldots . .+x^{2016}+\frac{x^{2017}}{1+x} \ldots(ii)
\)
Subtracting (i) from (ii)
\(
\begin{aligned}
& \frac{S}{1+x}=(1+x)^{2016}-\frac{x^{2017}}{1+x} \\
\therefore \quad & S=(1+x)^{2017}-x^{2017} \\
& a_{17}=\text { coefficient of } x^{17}={ }^{2017} C_{17}=\frac{2017!}{17!2000!}
\end{aligned}
\)
The sum of first 9 terms of the series.
\(
\frac{1^3}{1}+\frac{1^3+2^3}{1+3}+\frac{1^3+2^3+3^3}{1+3+5}+\ldots .
\) [JEE Main 2015]
\(
\begin{aligned}
& n ^{\text {th }} \text { term of series }=\frac{\left[\frac{ n ( n +1)}{2}\right]^2}{ n ^2}=\frac{1}{4}( n +1)^2 \\
& \text { Sum of } n \text { term }=\Sigma \frac{1}{4}( n +1)^2
\end{aligned}
\)
\(
=\frac{1}{4}\left[\Sigma n ^2+2 \Sigma n + n \right]
\)
\(
=\frac{1}{4}\left[\frac{ n ( n +1)(2 n +1)}{6}+\frac{2 n ( n +1)}{2}+ n \right]
\)
Sum of 9 terms
\(
=\frac{1}{4}\left[\frac{9 \times 10 \times 19}{6}+\frac{18 \times 10}{2}+9\right]=\frac{384}{4}=96
\)
If \(\sum_{n=1}^5 \frac{1}{n(n+1)(n+2)(n+3)}=\frac{k}{3}\), then \(k\) is equal to [Online April 11, 2015]
General term of given expression can be written as
\(
T_r=\frac{1}{3}\left[\frac{1}{n(n+1)(n+2)}-\frac{1}{(n+1)(n+2)(n+3)}\right]
\)
on taking summation both the side, we get
\(
\begin{aligned}
& \sum_{ r =1}^5 T _{ r }=\frac{1}{3}\left[\frac{1}{6}-\frac{1}{6.7 .8}\right]=\frac{ k }{3} \\
& \Rightarrow \frac{1}{3} \times \frac{1}{6}\left(1-\frac{1}{56}\right)=\frac{ k }{3} \Rightarrow \frac{1}{3} \times \frac{1}{6} \times \frac{55}{56}=\frac{ k }{3} \\
& \Rightarrow k =\frac{55}{336}
\end{aligned}
\)
The value of \(\sum_{r=16}^{30}(r+2)(r-3)\) is equal to : [Online April 10, 2015]
\(
\begin{aligned}
& \sum_{r=16}^{30}(r+2)(r-3) \\
& =\sum_{r=16}^{30}\left(r^2-r-6\right) \\
& =\sum_{r=1}^{30}\left(r^2-r-6\right)-\sum_{r=1}^{15}\left(r^2-r-6\right) \\
& =\left(\frac{30 \times 31 \times 61}{6}-\frac{30 \times 31}{2}-6 \times 30\right)-\left(\frac{15 \times 16 \times 31}{6}-\frac{15 \times 16}{2}-6 \times 15\right) \\
& =8810-1030 \\
& =7780
\end{aligned}
\)
\(
\text { If }(10)^9+2(11)^1\left(10^8\right)+3(11)^2(10)^7+\ldots . .
\)
\(+10(11)^9=k(10)^9\), then \(k\) is equal to: [JEE Main 2014]
\(
\begin{aligned}
& \text { Let } 10^9+2 \cdot(11)(10)^8+3(11)^2(10)^7 \\
& \quad+\ldots+10(11)^9=k(10)^9 \\
& \text { Let } x=10^9+2 \cdot(11)(10)^8+3(11)^2(10)^7+\ldots+10(11)^9
\end{aligned}
\)
Multiplied by \(\frac{11}{10}\) on both the sides
\(
\begin{aligned}
& \frac{11}{10} x=11 \cdot 10^8+2 \cdot(11)^2 \cdot(10)^7+\ldots+9(11)^9+11^{10} \\
& x\left(1-\frac{11}{10}\right)=10^9+11(10)^8+11^2 \times(10)^7 \\
& \quad+\ldots+11^9-11^{10}
\end{aligned}
\)
\(
\begin{aligned}
& -\frac{x}{10}=10^9\left[\frac{\left(\frac{11}{10}\right)^{10}-1}{\frac{11}{10}-1}\right]-11^{10} \\
& -\frac{x}{10}=\left(11^{10}-10^{10}\right)-11^{10}=-10^{10} \\
& x=10^{11}= k \cdot 10^9 \text { Given } \\
& k=100
\end{aligned}
\)
The number of terms in an A.P. is even; the sum of the odd terms in it is 24 and that the even terms is 30 . If the last term exceeds the first term by \(10 \frac{1}{2}\), then the number of terms in the A.P. is: [Online April 19, 2014]
Let \(a, d\) and \(2 n\) be the first term, common difference and total number of terms of an A.P. respectively i.e.
\(
a+(a+d)+(a+2 d)+\ldots+(a+(2 n-1) d)
\)
No. of even terms \(=n\), No. of odd terms \(=n\) Sum of odd terms :
\(
\begin{aligned}
& S _{ o }=\frac{n}{2}[2 a+(n-1)(2 d)]=24 \\
& \Rightarrow n[a+(n-1) d]=24 \dots(1)
\end{aligned}
\)
Sum of even terms:
\(
\begin{aligned}
& S _{ e }=\frac{n}{2}[2(a+d)+(n-1) 2 d]=30 \\
& \Rightarrow n[a+d+(n-1) d]=30 \dots(2)
\end{aligned}
\)
Subtracting equation (1) from (2), we get
\(
n d=6 \dots(3)
\)
Also, given that last term exceeds the first term by \(\frac{21}{2}\)
\(
\begin{aligned}
& a+(2 n-1) d=a+\frac{21}{2} \\
& 2 n d-d=\frac{21}{2} \\
& \Rightarrow 2 \times 6-\frac{21}{2}=d \\
& (\because n d=6) \\
& d=\frac{3}{2} \\
&
\end{aligned}
\)
Putting value of \(d\) in equation (3)
\(
n=\frac{6 \times 2}{3}=4
\)
Total no. of terms \(=2 n=2 \times 4=8\)
If the sum \(\frac{3}{1^2}+\frac{5}{1^2+2^2}+\frac{7}{1^2+2^2+3^2}+\ldots \ldots+\) up to 20 terms is equal to \(\frac{ k }{21}\), then \(k\) is equal to: [Online April 9, 2014]
\(n^{\text {th }}\) term of given series is
\(
\frac{2 n+1}{\frac{n(n+1)(2 n+1)}{6}}=\frac{6}{n(n+1)}
\)
Let \(n^{\text {th }}\) term, \(a_n=6\left[\frac{1}{n}-\frac{1}{n+1}\right]\)
Sum of 20 terms, \(S _{20}=a_1+a_2+a_3+\ldots .+a_{20}\)
\(
\begin{aligned}
S _{20}=6\left(\frac{1}{1}\right. & \left.-\frac{1}{2}\right)+6\left(\frac{1}{2}-\frac{1}{3}\right)+6\left(\frac{1}{3}-\frac{1}{4}\right)+\ldots \\
& +6\left(\frac{1}{18}-\frac{1}{19}\right)+6\left(\frac{1}{19}-\frac{1}{20}\right)+6\left(\frac{1}{20}-\frac{1}{21}\right)
\end{aligned}
\)
\(
S _{20}=6\left(1-\frac{1}{21}\right)=\frac{120}{21} \dots(2)
\)
Given that \(S _{20}=\frac{k}{21}\)
On comparing (1) and (2), we get
\(
k=120
\)
The sum of first 20 terms of the sequence \(0.7,0.77,0.777, \ldots .\). , is [JEE Main 2013]
Given sequence can be written as
\(
\begin{aligned}
& \frac{7}{10}+\frac{77}{100}+\frac{777}{10^3}+\ldots .+ \text { up to } 20 \text { terms } \\
& =7\left[\frac{1}{10}+\frac{11}{100}+\frac{111}{10^3}+\ldots . .+ \text { up to } 20 \text { terms }\right]
\end{aligned}
\)
Multiply and divide by 9
\(
\begin{array}{r}
=\frac{7}{9}\left[\frac{9}{10}+\frac{99}{100}+\frac{999}{1000}+\ldots . . \text { up to } 20 \text { terms }\right] \\
=\frac{7}{9}\left[\left(1-\frac{1}{10}\right)+\left(1-\frac{1}{10^2}\right)+\left(1-\frac{1}{10^3}\right)\right] \\
+\ldots \ldots \text { up to } 20 \text { terms }]
\end{array}
\)
\(
\begin{aligned}
& =\frac{7}{9}\left[20-\frac{\frac{1}{10}\left(1-\left(\frac{1}{10}\right)^{20}\right)}{1-\frac{1}{10}}\right] \\
& =\frac{7}{9}\left[\frac{179}{9}+\frac{1}{9}\left(\frac{1}{10}\right)^{20}\right] \\
& =\frac{7}{81}\left[179+(10)^{-20}\right]
\end{aligned}
\)
The value of \(1^2+3^2+5^2+\) \(\ldots\) \(+25^2\) is: [Online April 25, 2013]
Consider \(1^2+3^2+5^2+\ldots . . .+25^2\)
\(n^{\text {th }}\) term \(T _n=(2 n-1)^2, n=1, \ldots . .13\)
Now, \(S _n=\sum_{n=1}^{13} T _n=\sum_{n=1}^{13}(2 n-1)^2\)
\(
\begin{aligned}
& =\sum_{n=1}^{13} 4 n^2+\sum_{n=1}^{13} 1-\sum_{n=1}^{13} 4 n \\
& =4 \sum n^2+13-4 \sum n \\
& =4\left[\frac{n(n+1)(2 n+1)}{6}\right]+13-4 \frac{n(n+1)}{2}
\end{aligned}
\)
Put \(n=13\), we get
\(
\begin{aligned}
S _n & =26 \times 14 \times 9+13-26 \times 14 \\
& =3276+13-364 \\
& =2925 .
\end{aligned}
\)
The sum of the series : \((2)^2+2(4)^2+3(6)^2+\ldots\) upto 10 terms is: [Online April 23, 2013]
\(
\begin{aligned}
& 2^2+2(4)^2+3(6)^2+\ldots . . \text { upto } 10 \text { terms } \\
& =2^2\left[1^3+2^3+3^3+\ldots . . \text { upto } 10 \text { terms }\right] \\
& =4 \cdot\left(\frac{10 \times 11}{2}\right)^2=12100
\end{aligned}
\)
The sum \(\frac{3}{1^2}+\frac{5}{1^2+2^2}+\frac{7}{1^2+2^2+3^2}+\ldots\). upto 11-terms is: [Online April 22, 2013]
Given sum is
\(
\begin{aligned}
& \frac{3}{12}+\frac{5}{1^2+2^2}+\frac{7}{1^2+2^2+3^2}+\ldots . \\
& n \text {th term }= T _n \\
& =\frac{2 n+1}{\frac{n(n+1)(2 n+1)}{6}}=\frac{6}{n(n+1)} \\
& \text { or } T _n=6\left[\frac{1}{n}-\frac{1}{n+1}\right] \\
& S _n=\sum T _n=6 \sum \frac{1}{n}-6 \sum \frac{1}{n+1}=\frac{6 n}{n}-\frac{6}{n+1} \\
& =6-\frac{6}{n+1}=\frac{6 n}{n+1}
\end{aligned}
\)
So, sum upto 11 terms means
\(
S _{11}=\frac{6 \times 11}{11+1}=\frac{66}{12}=\frac{33}{6}=\frac{11}{2}
\)
The sum of the series : \(1+\frac{1}{1+2}+\frac{1}{1+2+3}+\ldots . .\). upto 10 terms, is : [Online April 9, 2013]
\(
\begin{aligned}
T _r & =\frac{1}{1+2+3+\ldots+r}=\frac{2}{r(r+1)} \\
& S _{10}=2 \sum_{r=1}^{10} \frac{1}{r(r+1)}=2 \sum_{r=1}^{10}\left[\frac{r+1}{r(r+1)}-\frac{r}{r(r+1)}\right] \\
& =2 \sum_{r=1}^{10}\left(\frac{1}{r}-\frac{1}{r+1}\right) \\
& =2\left[\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\ldots+\left(\frac{1}{10}-\frac{1}{11}\right)\right] \\
& =2\left[1-\frac{1}{11}\right]=2 \times \frac{10}{11}=\frac{20}{11}
\end{aligned}
\)
Statement-1: The sum of the series \(1+(1+2+4)+\) \((4+6+9)+(9+12+16)+\ldots .+(361+380+400)\) is 8000 .
Statement-2: \(\sum_{k=1}^n\left(k^3-(k-1)^3\right)=n^3\), for any natural number \(n\). [JEE Main 2012]
\(n\)th term of the given series
\(
\begin{aligned}
& =T_n=(n-1)^2+(n-1) n+n^2 \\
& =\frac{\left((n-1)^3-n^3\right)}{(n-1)-n}=n^3-(n-1)^3 \\
& \Rightarrow S_n=\sum_{k=1}^n\left[k^3-(k-1)^3\right] \Rightarrow 8000=n^3
\end{aligned}
\)
\(\Rightarrow n=20\) which is a natural number.
Now, put \(n=1,2,3, \ldots .20\)
\(
\begin{aligned}
& T_1=1^3-0^3 \\
& T_2=2^3-1^3
\end{aligned}
\)
\(
T_{20}=20^3-19^3
\)
Now, \(T_1+T_2+\cdots+T_{20}=S_{20}\)
\(
\Rightarrow \quad S_{20}=20^3-0^3=8000
\)
Hence, both the given statement is true.
If the sum of the series \(1^2+2.2^2+3^2+2.4^2+5^2+\ldots 2.6^2+\ldots\) upto \(n\) terms, when \(n\) is even, is \(\frac{n(n+1)^2}{2}\), then the sum of the series, when \(n\) is odd, is [Online May 26, 2012]
If \(n\) is odd, the required sum is
\(
\begin{aligned}
& 1^2+2.2^2+3^2+2.4^2+\ldots .+2(n-1)^2+n^2 \\
& =\frac{(n-1)(n-1+1)^2}{2}+n^2 \quad(\because n-1 \text { is even }) \\
& =\left(\frac{n-1}{2}+1\right) n^2=\frac{n^2( n +1)}{2}
\end{aligned}
\)
The sum of the series \(1+\frac{4}{3}+\frac{10}{9}+\frac{28}{27}+\ldots\) upto \(n\) terms is [Online May 19, 2012]
Given series is \(1+\frac{4}{3}+\frac{10}{9}+\frac{28}{27}+\ldots . n\) terms
\(
\begin{aligned}
& =1+\left(1+\frac{1}{3}\right)+\left(1+\frac{1}{9}\right)+\left(1+\frac{1}{27}\right)+\ldots n \text { terms } \\
& =(1+1+1+\ldots+n \text { terms }) \\
& +\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\ldots . n \text { terms }\right) \\
& =n+\frac{\frac{1}{3}\left(1-\frac{1}{3^n}\right)}{1-\frac{1}{3}}=n+\frac{1}{3} \times \frac{3}{2}\left[1-3^{-n}\right] \\
& =n+\frac{1}{2}\left[1-3^{-n}\right]=n+\frac{1}{2}-\frac{1}{2.3^n}
\end{aligned}
\)
The sum of the series
\(
\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\ldots
\)
upto 15 terms is [Online May 12, 2012]
Given series is
\(
\begin{aligned}
& \frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\ldots . . \\
& n^{\text {th }} \text { term }=\frac{1}{\sqrt{n}+\sqrt{n+1}}
\end{aligned}
\)
\(
\therefore 15^{\text {th }} \text { term }=\frac{1}{\sqrt{15}+\sqrt{16}}
\)
Thus, given series upto 15 terms is
\(
\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\ldots . .+\frac{1}{\sqrt{15}+\sqrt{16}}
\)
This can be re-written as
\(\frac{1-\sqrt{2}}{-1}+\frac{\sqrt{2}-\sqrt{3}}{-1}+\frac{\sqrt{3}-\sqrt{4}}{-1}+\ldots . .+\frac{\sqrt{15}-\sqrt{16}}{-1}\) (By rationalization)
\(
\begin{array}{r}
=-1+\sqrt{2}-\sqrt{2}+\sqrt{3}-\sqrt{3}+\sqrt{4}+\ldots . . \sqrt{14}+\sqrt{15} \\
-\sqrt{15}+\sqrt{16}
\end{array}
\)
\(
=-1+\sqrt{16}=-1+4=3
\)
Hence, the required sum \(=3\)
The sum of the series \(1^2+2.2^2+3^2+2.4^2+5^2+2.6^2+\ldots .+2(2 m)^2\) is [Online May 7, 2012]
The sum of the given series \(1^2+2.2^2+3^2+2.4^2+5^2+[latex] [latex]2.6^2+\ldots \ldots . .+2(2 m)^2\) is \(\frac{2 m(2 m+1)^2}{2}=m(2 m+1)^2\)
The sum to infinite term of the series \(1+\frac{2}{3}+\frac{6}{3^2}+\frac{10}{3^3}+\frac{14}{3^4}+\cdots\) is [JEE Main 2009]
We have
\(
S=1+\frac{2}{3}+\frac{6}{3^2}+\frac{10}{3^3}+\frac{14}{3^4}+\ldots \ldots \infty \dots(1)
\)
Multiplying both sides by \(\frac{1}{3}\) we get
\(
\frac{1}{3} S=\frac{1}{3}+\frac{2}{3^2}+\frac{6}{3^3}+\frac{10}{3^4}+\ldots \ldots \ldots \infty \dots(2)
\)
Subtracting eqn. (2) from eqn. (1) we get
\(
\begin{aligned}
&\begin{aligned}
& \frac{2}{3} S=1+\frac{1}{3}+\frac{4}{3^2}+\frac{4}{3^3}+\frac{4}{3^4}+\ldots \ldots \infty \\
& \frac{2}{3} S=\frac{4}{3}+\frac{4}{3^2}+\frac{4}{3^3}+\frac{4}{3^4}+\ldots \ldots \infty
\end{aligned}\\
&\frac{2}{3} S=\frac{\frac{4}{3}}{1-\frac{1}{3}}=\frac{4}{3} \times \frac{3}{2} \Rightarrow S=3
\end{aligned}
\)
The sum of series \(\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\) \(\ldots\) upto infinity is [JEE Main 2007]
\(
\text { We know that } e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\ldots \ldots . . \infty
\)
Put \(x=-1[latex]
[latex]
\begin{aligned}
& \therefore \quad e ^{-1}=1-1+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!} \ldots \ldots \infty \\
& \therefore \quad e ^{-1}=\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!} \ldots \ldots \infty
\end{aligned}
\)
The sum of the series \(1+\frac{1}{4.2!}+\frac{1}{16.4!}+\frac{1}{64.6!}+\ldots \ldots \ldots \ldots \ldots . . . . . . . . . . . . \text { upto infinity is }\) [JEE Main 2005]
\(
\begin{aligned}
& \frac{e^x+e^{-x}}{2}=1+\frac{x^2}{2!}+\frac{x^4}{4!}+\frac{x^6}{6!} \ldots \ldots \ldots . . \\
& \text { Putting } x=\frac{1}{2} \text { we get } \\
& 1+\frac{1}{4.2!}+\frac{1}{16.4!}+\frac{1}{64.6!}+\ldots \ldots \\
& \infty=\frac{e^{\frac{1}{2}}+e^{\frac{-1}{2}}}{2}=\frac{\sqrt{e}+\frac{1}{\sqrt{e}}}{2}=\frac{e+1}{2 \sqrt{e}}
\end{aligned}
\)
The sum of series \(\frac{1}{2!}+\frac{1}{4!}+\frac{1}{6!}+\ldots .\). is [JEE Main 2004]
We know that \(e=1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\ldots \ldots\)
and \(e^{-1}=1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\ldots \ldots\)
\(
\begin{aligned}
& \therefore e+e^{-1}=2\left[1+\frac{1}{2!}+\frac{1}{4!}+\ldots .\right] \\
& \therefore \frac{1}{2!}+\frac{1}{4!}+\frac{1}{6!}+\ldots \ldots=\frac{e+e^{-1}}{2}-1 \\
& =\frac{e^2+1-2 e}{2 e}=\frac{(e-1)^2}{2 e}
\end{aligned}
\)
The sum of the first \(n\) terms of the series
\(
1^2+2.2^2+3^2+2.4^2+5^2+2.6^2+\ldots
\)
is \(\frac{n(n+1)^2}{2}\) when \(n\) is even. When \(n\) is odd the sum is [JEE Main 2004]
If \(n\) is odd, the required sum is
\(
\begin{aligned}
& 1^2+2.2^2+3^2+2.4^2+\ldots . .+2 \cdot(n-1)^2+n^2 \\
& =\frac{(n-1)(n-1+1)^2}{2}+n^2 \\
& {[\because(n-1) \text { is even }}
\end{aligned}
\)
\(\therefore\) using given formula for the sum of \((n-1)\) terms.]
\(
=\left(\frac{n-1}{2}+1\right) n^2=\frac{n^2(n+1)}{2}
\)
If \(S_n=\sum_{r=0}^n \frac{1}{{ }^n C_r}[latex] and [latex]t_n=\sum_{r=0}^n \frac{r}{{ }^n C_r}\), then \(\frac{t_n}{S_n}\) is equal to [JEE Main 2004]
\(
\begin{aligned}
S_n & =\frac{1}{{ }^n C_0}+\frac{1}{{ }^n C_1}+\frac{1}{{ }^n C_2}+\ldots .+\frac{1}{{ }^n C_n} \\
t_n & =\frac{0}{{ }^n C_0}+\frac{1}{{ }^n C_1}+\frac{2}{{ }^n C_2}+\ldots .+\frac{n}{{ }^n C_n} \\
t_n & =\frac{n}{{ }^n C_n}+\frac{n-1}{{ }^n C_{n-1}}+\frac{n-2}{{ }^n C_{n-2}}+\ldots .+\frac{0}{{ }^n C_0}
\end{aligned}
\)
Add, \(2 t_n=(n)\left[\frac{1}{{ }^n C_0}+\frac{1}{{ }^n C_1}+\ldots . \frac{1}{{ }^n C_n}\right]=n S_n\)
\(
\therefore \frac{t_n}{S_n}=\frac{n}{2}
\)
The sum of the series \(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{3.4} \ldots \ldots . .\). up to \(\infty\) is equal to [JEE Main 2003]
\(
\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{3.4} \ldots \ldots \ldots \ldots \infty
\)
\(
\begin{aligned}
& \left|T_n\right|=\frac{1}{n(n+1)}=\left(\frac{1}{n}-\frac{1}{n+1}\right) \\
& S=T_1-T_2+T_3-T_4+T_5 \ldots \ldots \ldots \infty
\end{aligned}
\)
\(
=\left(\frac{1}{1}-\frac{1}{2}\right)-\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)-\left(\frac{1}{4}-\frac{1}{5}\right) \ldots . .
\)
\(
=1-2\left[\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5} \ldots \ldots \ldots \ldots \ldots \infty\right]
\)
\(
=1-2[-\log (1+1)+1]=2 \log 2-1=\log \left(\frac{4}{e}\right)
\)
\(1^3-2^3+3^3-4^3+\ldots+9^3=\) [JEE Main 2002]
\(
\begin{aligned}
& 1^3-2^3+3^3-4^3+\ldots \ldots .+9^3 \\
& =1^3+2^3+3^3+\ldots \ldots .+9^3-2\left(2^3+4^3+6^3+8^3\right) \\
& =\left[\frac{9 \times 10}{2}\right]^2-2.2^3\left[1^3+2^3+3^3+4^3\right] \\
& =(45)^2-16 \cdot\left[\frac{4 \times 5}{2}\right]^2=2025-1600=425
\end{aligned}
\)
The value of \(2^{1 / 4} \cdot 4^{1 / 8} \cdot 8^{1 / 16} \ldots \infty\) is [JEE Main 2002]
The product is \(P=2^{1 / 4} \cdot 2^{2 / 8} \cdot 2^{3 / 16}\) \(\ldots\) \(=2^{1 / 4+2 / 8+3 / 16+\ldots \ldots \ldots \infty}\)
\(
\text { Now let } S=\frac{1}{4}+\frac{2}{8}+\frac{3}{16}+\ldots \ldots . \infty \dots(1)
\)
\(
\frac{1}{2} S=\frac{1}{8}+\frac{2}{16}+\ldots \ldots \infty \dots(2)
\)
Subtracting (2) from (1)
\(
\Rightarrow \quad \frac{1}{2} S=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\ldots \ldots . . \infty
\)
\(
\begin{aligned}
& \text { or } \quad \frac{1}{2} S=\frac{1 / 4}{1-1 / 2}=\frac{1}{2} \Rightarrow S=1 \\
& \therefore P=2^S=2
\end{aligned}
\)
You cannot copy content of this page