JEE Main 2025 — Sequences And Series Question with Solution

To solve this problem, we start by analyzing the terms of an arithmetic progression (A.P.) where:

$T_m=\frac{1}{25}$

$T_{25}=\frac{1}{20}$

$20\sum _{r=1}^{25}{T}_r=13$

The formula for the $r^{\text{th}}$ term of an A.P. is:

$T_r=a+(r-1)d$

Given:

$T_m=a+(m-1)d=\frac{1}{25}\text{(Equation 1)}$

$T_{25}=a+24d=\frac{1}{20}$

The sum of the first 25 terms ($S_{25}$) is given by:

$S_{25}=\frac{25}{2}\times (2a+24d)$

Substituting into the equation for the sum:

$20\times \frac{25}{2}\times (2a+24d)=13$

This simplifies to:

$a=\frac{1}{500}$

Substituting $a=\frac{1}{500}$ into $T_{25}=a+24d=\frac{1}{20}$, we find:

$\frac{1}{500}+24d=\frac{1}{20}$

$24d=\frac{1}{20}-\frac{1}{500}$

$d=\frac{1}{500}$

Using Equation 1 again:

$\frac{1}{500}+\frac{m-1}{500}=\frac{1}{25}$

$\frac{m}{500}=\frac{1}{25}$

$m=20$

Now to find $5m\sum _{r=m}^{2m}{T}_r$:

$5m=5\times 20=100$

$\sum _{r=20}^{40}{T}_r=\frac{21}{2}\times (T_{20}+T_{40})$

Since $m=20$, the summation covers terms from $T_{20}$ to $T_{40}$, and:

$100\times \sum _{r=20}^{40}{T}_r=126$

Therefore, $5m\sum _{r=m}^{2m}{T}_r=126$.