Let Tr be the r^th term of an A.P. If for some m, ${\mathrm{T}}_{\mathrm{m}}=\frac{1}{25},{\mathrm{T}}_{25}=\frac{1}{20}\text{ and }20\sum _{\mathrm{r}=1}^{25} {\mathrm{T}}_{\mathrm{r}}=13,$ then $5\mathrm{m}\sum _{\mathrm{r}=\mathrm{m}}^{2\mathrm{m}} {\mathrm{T}}_{\mathrm{r}}$ is equal to :

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

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) $\frac{1}{500}+\frac{m-1}{500}=\frac{1}{25}\Rightarrow m=20$

Now,

$5\mathrm{m}\sum _{\mathrm{r}=\mathrm{m}}^{2\mathrm{m}} {\mathrm{T}}_{\mathrm{r}}=100\sum _{\mathrm{r}=20}^{40} {\mathrm{T}}_{\mathrm{r}}=126$