JEE Main 2026 — Mathematics Algebra

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

If the sum of the first four terms of an A.P. is 6 and the sum of its first six terms is 4, then the sum of its first twelve terms is