Let $$T_{r}$$ be the $$r^{th}$$ term of an A.P. If for some m,$$T_{m}=\frac{1}{25},T_{25}=\frac{1}{25}$$, and $$20\sum_{r=1}^{25}T_{r}=13$$,then $$5m\sum_{r=m}^{2m}T_{r}$$ is equal to

Standard AP property: If $$T_m = \frac{1}{n}$$ and $$T_n = \frac{1}{m}$$, then common difference $$d = \frac{1}{mn}$$ and first term $$a = \frac{1}{mn}$$.

Here, $$n=25$$, so $$a = d = \frac{1}{25m}$$.

Sum formula: $$S_{25} = \frac{25}{2}[2a + 24d] = \frac{25}{2}[26d] = 25 \cdot 13d = \frac{25 \cdot 13}{25m} = \frac{13}{m}$$.

Given $$20(S_{25}) = 13 \implies 20(\frac{13}{m}) = 13 \implies m = 20$$.

Now, $$a = d = \frac{1}{500}$$. We need $$5m(S_{2m} - S_{m-1}) = 100(S_{40} - S_{19})$$.

$$S_k = \frac{k}{2}[2d + (k-1)d] = \frac{k(k+1)d}{2}$$

$$100 \left[ \frac{40 \cdot 41}{2 \cdot 500} - \frac{19 \cdot 20}{2 \cdot 500} \right] = 100 \left[ \frac{1640 - 380}{1000} \right] = \frac{1260}{10} = \mathbf{126}$$