$VT=\frac{1}{3}.(\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{n(n+3)})$$=\frac{1}{3}.(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{n}-\frac{1}{n+3})$
$=\frac{1}{3}(\frac{1}{5}-\frac{1}{n+3})=\frac{1}{20}$
$\Rightarrow \frac{1}{5}-\frac{1}{n+3}=\frac{3}{20}$
$\Rightarrow \frac{1}{n+3}=\frac{1}{20}\Rightarrow n=17$