Question

Prove that the maximum value of the sum of the series $20+19\frac{1}{3}+18\frac{2}{3}+$______ is $310$.

Answer 1

Given series is an A.P. of common difference $-\frac{2}{3}.$
Hence the sum will be maximum when all the terms taken are positive. $T_n=20+(n-1)(-23)\ge 0$ if $62-2n>0$ or $n\le 31$
$\therefore S_{31}$ is max.
Hence $S_{31}=\frac{31}{2}\left(2\times 20+30(-\frac{2}{3})\right)=\frac{31}{2}\times \left(40-20\right)=310.$