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

The given series is arithmetic whose first term = 20, common difference = – . As the common difference is negative, the terms  will become negative after some stage. So the sum is maximum if only positive terms are added.

Now, $\begin{array}{l}{\mathrm{t}}_{\mathrm{n}}=20+(\mathrm{n}-1)\left(-\frac{2}{3}\right)\ge 0\text{ if }60-2(\mathrm{n}-1)\ge 0\\ \text{ or }62\ge 2\mathrm{n}\text{ or }31\ge \mathrm{n}\end{array}$

$\therefore$ The first 31 terms are non-negative.

$\begin{array}{l}\therefore \mathrm{maximum} \mathrm{sum}={\mathrm{S}}_{31}=\frac{31}{2}\left[2\times 20+(31-1)\left(-\frac{2}{3}\right)\right]\\ =\frac{31}{2}(40-20)-310\end{array}$