Question:
In the sum of the series
$20+19 \frac{3}{5}+19 \frac{1}{5}+18 \frac{4}{5}+\ldots$ upto $n^{\text {th }}$ term is 488 and then
$n^{\text {th }}$ term is negative, then :
Correct Option: , 2
Solution:
$S_{n}=20+19 \frac{3}{5}+19 \frac{1}{5}+18 \frac{4}{5}+\ldots$
$\because S_{n}=488$
$488=\frac{n}{2}\left[2\left(\frac{100}{5}\right)+(n-1)\left(-\frac{2}{5}\right)\right]$
$488=\frac{n}{2}(101-n) \Rightarrow n^{2}-101 n+2440=0$
$\Rightarrow n=61$ or 40
For $n=40 \Rightarrow T_{n}>0$
For $n=61 \Rightarrow T_{n}<0$
$n^{\text {th }}$ term $=T_{61}=\frac{100}{5}+(61-1)\left(-\frac{2}{5}\right)=-4$