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Q.
The maximum sum of the series $20+19 \frac{1}{3}+18 \frac{2}{3}+\cdots$ is
Sequences and Series
Solution:
$n$th term of the series is $20+(n-1)\left(-\frac{2}{3}\right)$. For sum to be maximum,
$n$th term $\geq 0 \Rightarrow 20+(n-1)\left(-\frac{2}{3}\right) \geq 0 \Rightarrow n \leq 31$
Thus, the sum of $31$ terms is maximum and is equal to
$\frac{31}{2}\left[40+30 \times\left(-\frac{2}{3}\right)\right]=310$