Let the first term be $a$ and the common difference be $d$, then

 $20=\frac{1}{a+4d}$ and $50=\frac{1}{a+5d}$
On solving, we get,

 $d=-\frac{3}{100}$ and $a=\frac{17}{100}$
Hence, $t_r=\frac{100}{20-3r}>0\Rightarrow r<\frac{20}{3}\Rightarrow r_{\mathrm{m}\mathrm{a}\mathrm{x}}=6$
So, sum is maximum for $n=\mathrm{6,}$ as after this terms are negative