Question

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

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

Answer 1

The correct option is A

310

$n^{th}$ term of the series is 20+(n-1) (-2/3).
For the sum to be maximum,
$n^{th}$ term $\ge$0
$\Rightarrow 20+(n-1)(-\frac{2}{3})\ge 0$

$\Rightarrow n\le 31$
Thus, the sum of 31 terms is maximum and is

$\frac{31}{2}[40+30\times (-\frac{2}{3})]=310$