Given, if  $S=\frac{7}{5}+\frac{9}{5^2}+\frac{13}{5^3}+\frac{19}{5^4}+\dots \dots \infty ...\left(i\right)$

Dividing by $5$ on both sides of series

$\frac{S}{5}=\frac{7}{5^2}+\frac{9}{5^3}+\frac{13}{5^4}+\frac{19}{5^5}+\dots \dots \infty ...\left(ii\right)$ 

Subtracting equations  $\left(i\right)\&\left(ii\right)$

$S-\frac{S}{5}=\left[\frac{7}{5}+\frac{9}{5^2}+\frac{13}{5^3}+\frac{19}{5^4}+\dots \dots \infty \right]-\left[\frac{7}{5^2}+\frac{9}{5^3}+\frac{13}{5^4}+\frac{19}{5^5}+\dots \dots \infty \right]$

$\frac{4S}{5}=\left[\frac{7}{5}+\left(\frac{9}{5^2}-\frac{7}{5^2}\right)+\left(\frac{13}{5^3}-\frac{9}{5^3}\right)+\left(\frac{19}{5^4}-\frac{13}{5^4}\right)+\dots \dots \infty \right]$

$\frac{4S}{5}=\frac{7}{5}+\frac{2}{5^2}+\frac{4}{5^3}+\frac{6}{5^4}+\frac{8}{5^5}+\dots \dots \infty$

$\left(\frac{4S}{5}-\frac{7}{5}\right)=k=\frac{2}{5^2}+\frac{4}{5^3}+\frac{6}{5^4}+\frac{8}{5^5}+\dots ..\infty ...\left(iii\right)$

Let $\left(\frac{4S}{5}-\frac{7}{5}\right)=k$

$\frac{k}{5}=\frac{2}{5^3}+\frac{4}{5^4}+\frac{6}{5^5}+\frac{8}{5^6}+\dots ..\infty ...\left(iv\right)$

Subtracting equation $\left(iii\right) \& \left(iv\right)$

$\frac{4k}{5}=\frac{2}{5^2}+\frac{2}{5^3}+\frac{2}{5^4}+\frac{2}{5^5}\dots \dots \infty$

Common ratio

$r=\frac{{\displaystyle \frac{2}{5^3}}}{{\displaystyle \frac{2}{5^2}}}=\frac{1}{5}, r=\frac{{\displaystyle \frac{2}{5^4}}}{{\displaystyle \frac{2}{5^3}}}=\frac{1}{5}$

$\frac{4k}{5}=\frac{2}{25}\left\{\frac{1}{1-\frac{1}{5}}\right\}=\frac{1}{10}$

$k=\frac{1}{8}$

 Now $\frac{4S}{5}-\frac{7}{5}=\frac{1}{8}$

$\frac{4S}{5}=\frac{7}{5}+\frac{1}{8}$

$S=\frac{61}{32}$

$160S=160\times \frac{61}{32}$

$=305$