Let for n=1,2,……,50,Sn be the sum of the infinite geometric progression whose first term is n2 and whose common ratio is (n+1)21. Then the value of 261+n=1∑50(Sn+n+12−n−1) is equal to
Held on 28 Jun 2022 · Verified 6 Jul 2026.
Sign in to track your attempts and accuracy.
Sign in to keep a private note on this question. Nothing you write is ever public.
Let $a_{1}=1$ and for $n \geqslant 1, a_{n+1}=\frac{1}{2} a_{n}+\frac{n^{2}-2 n-1}{n^{2}(n+1)^{2}}$. Then $\left|\sum_{n=1}^{\infty}\left(a_{n}-\frac{2}{n^{2}}\right)\right|$ is equal to $\_\_\_\_$.
Let $a_{1}, a_{2}, a_{3}, \ldots$ be G.P. of increasing positive terms such that $a_{2} \cdot a_{3} \cdot a_{4}=64$ and $a_{1}+a_{3}+a_{5}=\frac{813}{7}$. Then $a_{3}+a_{5}+a_{7}$ is equal to :
Let $\sum_{k=1}^{n} a_{k}=\alpha n^{2}+\beta n$. If $a_{10}=59$ and $a_{6}=7 a_{1}$, then $\alpha+\beta$ is equal to
Consider an A.P.: $a_{1}, a_{2}, \ldots, a_{\mathrm{n}} ; a_{1}>0$. If $a_{2}-a_{1}=\frac{-3}{4}, a_{\mathrm{n}}=\frac{1}{4} a_{1}$, and $\sum_{\mathrm{i}=1}^{\mathrm{n}} a_{\mathrm{i}}=\frac{525}{2}$, then $\sum_{\mathrm{i}=1}^{17} a_{\mathrm{i}}$ is equal to
Let $a_{1}, \frac{a_{2}}{2}, \frac{a_{3}}{2^{2}}, \ldots, \frac{a_{10}}{2^{9}}$ be a G.P. of common ratio $\frac{1}{\sqrt{2}}$. If $a_{1}+a_{2}+\ldots+a_{10}=62$, then $a_{1}$ is equal to :
Work through every JEE Main Algebra PYQ, year by year.