The maximum sum of the series $20+19\frac{1}{3}+18\frac{2}{3}+18+\mathrm{...........}$  is

The given series is arithmetic whose first term=20 and

 common difference $-\frac{2}{3}$

As the common difference is negative the terms will become negative after some stage. So the  sum is maximum when all positive terms are added now, for the positive terms

$x_n\ge 0\Rightarrow 20+\left(n-1\right)\times -\frac{2}{3}\ge 0$

$\Rightarrow 60-2\left(n-1\right)\ge 0\Rightarrow n\le 31$

$\therefore$  the first 31 terms are non-negative

$\therefore$ $S_{31}=\frac{31}{2}\left[2\times 20+\left(31-1\right)\times -\frac{2}{3}\right]=310$