Mathematics - Sequence & Series Question with Solution | TestHub

$\begin{array}{rl}& {\mathrm{T}}_{\mathrm{m}}=\frac{1}{25},{\mathrm{T}}_{25}=\frac{1}{20},20\sum _{\mathrm{r}=1}^{25}{\mathrm{T}}_{\mathrm{r}}=13\\ & {\mathrm{T}}_{\mathrm{m}}=\mathrm{a}+(\mathrm{m}-1)\mathrm{d}=\frac{1}{25}\dots \dots .(1)\\ & {\mathrm{T}}_{25}=\mathrm{a}+24\mathrm{d}=\frac{1}{20}\\ & 20\cdot \frac{25}{2}\left[\mathrm{a}+\frac{1}{20}\right]=13\Rightarrow \mathrm{a}=\frac{1}{500}\end{array}$ 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, $5\mathrm{m}{\sum }_{\mathrm{r}=\mathrm{m}}^{2\mathrm{m}}{\mathrm{T}}_{\mathrm{r}}=100{\sum }_{\mathrm{r}=20}^{40}{\mathrm{T}}_{\mathrm{r}}=126$