$$\begin{aligned} & \mathrm{T}_{\mathrm{m}}=\frac{1}{25}, \mathrm{~T}_{25}=\frac{1}{20}, 20 \sum_{\mathrm{r}=1}^{25} \mathrm{~T}_{\mathrm{r}}=13 \\ & \mathrm{~T}_{\mathrm{m}}=\mathrm{a}+(\mathrm{m}-1) \mathrm{d}=\frac{1}{25} \ldots \ldots .(1) \\ & \mathrm{T}_{25}=\mathrm{a}+24 \mathrm{~d}=\frac{1}{20} \end{aligned}$$

$20.\frac{25}{2}\left[a+\frac{1}{20}\right]=13 \Rightarrow a=\frac{1}{500}$

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,

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