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Q.
If the sum of the series $25+24 \frac{2}{5}+23 \frac{4}{5}+23 \frac{1}{5}+\ldots \ldots \ldots .$. is maximum, then
Sequences and Series
Solution:
Terms are $25,24 \frac{2}{5}, 23 \frac{4}{5}, 23 \frac{1}{5}, \ldots \ldots . .$.
which is in A.P. whose first term $=25$ and common difference $=\frac{-3}{5}$.
Now,
$T _{ n }=25+( n -1)\left(\frac{-3}{5}\right)<0 \Rightarrow n >42$
$\therefore(42)^{\text {nd }}$ term will be the last positive term.
So, the sum is maximum, when number of terms is 42 .