Question

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

• A. 310
• B. 300
• C. 320
• D. None of these

Answer 1

Answer: (A)

$n$th term of the series is $20+(n-1)\left(-\frac{2}{3}\right)$. For sum to be maximum,
$n$th term $\ge 0\Rightarrow 20+(n-1)\left(-\frac{2}{3}\right)\ge 0\Rightarrow n\le 31$
Thus, the sum of $31$ terms is maximum and is equal to
$\frac{31}{2}\left[40+30\times \left(-\frac{2}{3}\right)\right]=310$