JEE Main 2025 (Online) 28th January Morning ShiftQuestion 1
Let $T_r$ be the $r^{th}$ term of an A.P. If for some $m$, $T_m=\frac{1}{25}$, $T_{25}=\frac{1}{20}$, and $20\sum_{r=1}^{25}T_r=13$, then $5m\sum_{r=m}^{2m}T_r$ is equal to
Sequence and Series · Arithmetic Progression
Solution
Formulas used. (1) $S_n=\frac n2\,(T_1+T_n)$(2) $T_r=T_1+(r-1)d$Step 1 - First term. $$20\cdot\frac{25}{2}\left(T_1+T_{25}\right)=13\ \Longrightarrow\ T_1+\frac1{20}=\frac{13}{250}\ \Longrightarrow\ T_1=\frac{26-25}{500}=\frac1{500}.$$ Step 2 - Common difference. $$d=\frac{T_{25}-T_1}{24}=\frac{\frac{25}{500}-\frac1{500}}{24}=\frac1{500},\qquad\text{so } T_r=\frac{r}{500}.$$ Step 3 - Find $m$. $\dfrac m{500}=\dfrac1{25}$ gives $m=20$. Step 4 - The sum. $\sum_{r=20}^{40}r=\dfrac{21}{2}(20+40)=630$: $$5m\sum_{r=m}^{2m}T_r=100\cdot\frac{630}{500}=126 .$$ Answer: (C)