Question

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

• A. 300
• B. 310
• C. $311\frac{2}{3}$
• D. $333\frac{2}{3}$

Answer 1

The correct option is B 310

The given series is arthmetic 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
$x_n\ge 0\Rightarrow 20+(n-1)\times -\frac{2}{3}\ge 0\Rightarrow 60-2(n-1)\ge 0\Rightarrow n\le 31.$
$\therefore$ The first 31 terms are non-negative
$\therefore Maximumsum=s_{31}=\frac{31}{2}[2\times 20+(31-1)\times -\frac{2}{3}]=310$