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

The correct option is A $310$

$s=\frac{60}{3}+\frac{58}{3}+\frac{56}{3}+\frac{54}{3}+...$

$a=\frac{60}{3},d=\frac{58}{3}-\frac{60}{3}=-\frac{2}{3}$

$T_n=a+(n-1)d=0$

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

$\frac{60}{3}=(n-1)\frac{2}{3}$

$n=31$

$S_{31}=\frac{n}{2}(a+d)\Rightarrow \frac{31}{2}(\frac{60}{3}+0)$

$s_{31}=31\times 10=310$

$s_{31}=310$