JEE Main 2025 — Sequences And Series Question with Solution

To find the value of $S_{2025}$, calculate the partial sum

$S_n=\sum _{n=1}^{2025}\frac{1}{n(n+1)}=\sum _{n=1}^{2025}\left(\frac{1}{n}-\frac{1}{n+1}\right)$

This series is telescopic, meaning most terms cancel out with each other:

$=\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\cdots +\left(\frac{1}{2025}-\frac{1}{2026}\right)$

Simplifying the expression, we find:

$=\frac{1}{1}-\frac{1}{2026}=1-\frac{1}{2026}=\frac{2025}{2026}$

Now, evaluate $\sqrt{2026\cdot S_{2025}}$:

$\sqrt{2026\cdot \frac{2025}{2026}}=\sqrt{2025}=45$

For the arithmetic progression with the first term $-p$ and common difference $p$, the sum of the first six terms is given by:

$\frac{6}{2}[-2p+(6-1)p]=3(5p)=15p$

Given that:

$15p=45$

We find:

$p=3$

To determine the absolute difference between the $20^{\text{th}}$ and $15^{\text{th}}$ terms of the A.P., compute:

$|A_{20}-A_{15}|=|(-p+19p)-(-p+14p)|$

Substituting the value of $p$:

$=|18p-13p|=|5p|=5\times 5=25$

Thus, the absolute difference is $\boxed{25}$.