Given $\frac{1}{x_1},\frac{1}{x_2},\dots \dots \dots ,\frac{1}{x_n}$ are in A.P. and $x_1=4$ and $x_{21}=20$.

$\Rightarrow a=\frac{1}{4}$

Given, $x_{21}=20$.

$\Rightarrow \frac{1}{4}+20·d=\frac{1}{20}$

$\Rightarrow d=-\frac{1}{100}$

Given, $x_{\mathrm{n}}>\mathrm{50}$.

$\Rightarrow \frac{1}{x_n}<\frac{1}{50}$

$\Rightarrow \frac{1}{4}- \frac{n-1}{100}< \frac{1}{50}\Rightarrow n>24$

$\because n=25$

Now, $\sum _{i=1}^{25}\left(\frac{1}{x_i}\right)=\frac{25}{2}\left[2\times \frac{1}{4}-\frac{1}{100}\times 24\right]=\frac{13}{4}$