A) last term of the series is \[0\].
B) last term of the series is \[\frac{1}{5}\]
C) number of terms of the series are \[42\].
D) number of terms of the series are \[40\].
Correct Answer: C
Solution :
Terms are \[25,\,\,24\frac{2}{5},\,\,23\frac{4}{5},\,\,23\frac{1}{5},\,\,.........\] 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)nd term will be the last positive term. So, the sum is maximum, when number of terms is 42.
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