JEE Main 2025 — Sequences and Series Question with Solution

$\begin{aligned} & \mathrm{T}_{\mathrm{m}}=\frac{1}{25}, \mathrm{~T}_{25}=\frac{1}{20}, 20 \sum_{\mathrm{r}=1}^{25} \mathrm{~T}_{\mathrm{r}}=13 \\ & \mathrm{~T}_{\mathrm{m}}=\mathrm{a}+(\mathrm{m}-1) \mathrm{d}=\frac{1}{25} \ldots \ldots .(1) \\ & \mathrm{T}_{25}=\mathrm{a}+24 \mathrm{~d}=\frac{1}{20} \\ & 20 \cdot \frac{25}{2}\left[\mathrm{a}+\frac{1}{20}\right]=13 \Rightarrow \mathrm{a}=\frac{1}{500} \end{aligned}$ also, $20{\mathrm{S}}_{25}=20\cdot \frac{25}{2}[2\mathrm{a}+24\mathrm{d}]=13\Rightarrow \mathrm{d}=\frac{1}{500}$ from (1) $\frac{1}{500}+\frac{\mathrm{m}-1}{500}=\frac{1}{25}\Rightarrow \mathrm{m}=20$ Now, $5\mathrm{m}{\sum }_{\mathrm{r}=\mathrm{m}}^{2\mathrm{m}}{\mathrm{T}}_{\mathrm{r}}=100{\sum }_{\mathrm{r}=20}^{40}{\mathrm{T}}_{\mathrm{r}}=126$