Question

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

• A. $310$
• B. $290$
• C. $320$
• D. None of these

Answer 1

The correct option is A $310$

The given series is arithmetic whose first term $=20$, common difference $=-\frac{2}{3}$.

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 $t_n=20+(n-1)(-\frac{2}{3})\ge 0\Rightarrow 60-2(n-1)\ge 0$

$\Rightarrow 62\ge 2n\Rightarrow 31\ge n$

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

$\therefore$ Maximum sum $=S_{31}=\frac{31}{2}\{2\times 20+(31-1)(-\frac{2}{3})\}$

$=\frac{31}{2}(40-20)=310$