Let \(T_r\) be the \(r^{th}\) term of an A.P. If for some \(m,\) \({\mathrm{T}}_{\mathrm{m}}=\frac{1}{25},{\mathrm{T}}_{…

\({\text{Given, T}}_{r}\text{represents terms of A.P.}\\ {\mathrm{T}}_{\mathrm{m}}=\frac{1}{25},{\mathrm{T}}_{25}=\frac{1}{20},20\sum _{\mathrm{r}=1}^{25}{\mathrm{T}}_{\mathrm{r}}=13\\ 20\sum _{r=1}^{25}{T}_{r}=20\left[\frac{25}{2}\left[a+\frac{1}{20}\right]\right]=13\\ \Rightarrow a=\frac{1}{20\times 25}=\frac{1}{500}\\ ∵{T}_{25}=a+24d=\frac{1}{20}\\ \Rightarrow d=\frac{1}{20\times 25}=\frac{1}{500}\\ \text{Now},\\ {T}_{m}=a+(m-1)d=\frac{1}{25}\\ =\frac{1}{500}m=\frac{1}{25}\\ \Rightarrow m=20\\ \text{ Now, }5m\sum _{r=m}^{2m}{T}_{r}=5\times 20\left[\sum _{r=20}^{40}{T}_{r}\right]\\ =100\left[\frac{40}{2}(2a+39d)-\frac{19}{2}(2a+18d)\right]\\ \text{but a=d (we already find above)}\\ 5m\sum _{r=m}^{2m}{T}_{r}=100\left[\frac{40}{2}\times 41d-\frac{19}{2}\times 20d\right]\\ =126\)