$\begin{array}{l}\text{ A function }\mathrm{f}\left(\mathrm{x}\right)\text{ is given by }\mathrm{f}\left(\mathrm{x}\right)=\frac{5^{\mathrm{x}}}{5^{\mathrm{x}}+5}\text{ , then the sum of the series }\\ \mathrm{f}\left(\frac{1}{20}\right)+\mathrm{f}\left(\frac{2}{20}\right)+\mathrm{f}\left(\frac{3}{20}\right)+\dots \dots ..+\mathrm{f}\left(\frac{39}{20}\right)\text{ is equal to }\end{array}$

Given $f\left(x\right)=\frac{5^x}{5^x+5}$

$\begin{array}{c}f\left(x\right)+f(2-x)=\frac{5^x}{5^x+5}+\frac{5^{2-x}}{5^{2-x}+5}\\ =\frac{5^x}{5^x+5}+\frac{\frac{25}{5^x}}{\frac{25+5·5^x}{5^x}}\\ =\frac{5^x}{5^x+5}+\frac{5}{5^x+5}\\ =1\end{array}$

Consider 

$\begin{array}{c}f\left(\frac{1}{20}\right)+f\left(\frac{2}{20}\right)+\mathrm{...}+f\left(\frac{39}{20}\right)=f\left(\frac{1}{20}\right)+f\left(\frac{2}{20}\right)+\mathrm{...}+f\left(2-\frac{2}{20}\right)+f\left(2-\frac{1}{20}\right) ( club the first and last terem \& second and last but one term and ....etc)\\ =19\left(1\right)+f\left(\frac{20}{20}\right) =19 +f\left(1\right) =19+\frac{1}{2}\\ =\frac{39}{2}\end{array}$