Question

If the sum of the series $25+24\frac{2}{5}+23\frac{4}{5}+23\frac{1}{5}+\dots \dots \dots .$. is maximum, then

• 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 .

Answer 1

Answer: (C)

Terms are $25,24\frac{2}{5},23\frac{4}{5},23\frac{1}{5},\dots \dots ..$.
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 .