Let, $a \& d$ is the first term and common difference of arithmetic progression respectively, then

$T_{10}=\frac{1}{20}=a+9d ...\left(i\right)$
$T_{20}=\frac{1}{10}=a+19d ...\left(ii\right)$

After solving $\left(i\right)$ and $\left(ii\right)$, we get
$a=\frac{1}{200}, d=\frac{1}{200}$

Now

$S_{200}=\frac{200}{2}\left[\frac{2}{200}+\frac{199}{200}\right]=\frac{201}{2}=100\frac{1}{2}$.