Question

If the sum of the series $20+19\frac{3}{5}+19\frac{1}{5}+18\frac{4}{5}+\cdots$ upto $n^{th}$ term is $488$ and the $n^{th}$ term is negative, then:

• A. $n=60$
• B. $n=41$
• C. $n^{th}$ term is $-4$
• D. $n^{th}$ term is $-4\frac{2}{5}$

Answer 1

The correct option is C $n^{th}$ term is $-4$

$\frac{100}{5}+\frac{98}{5}+\frac{96}{5}+\cdots$ upto $n^{th}$ term
$S_n=488,a=20,d=-\frac{2}{5}$
$\Rightarrow \frac{n}{2}[2\times 20+(n-1)\left(\frac{-2}{5}\right)]=488$
$\Rightarrow 20n-\frac{n^2}{5}+\frac{n}{5}=488$
$\Rightarrow 100n-n^2+n=2440$
$\Rightarrow n^2-101n+2440=0$
$\Rightarrow n=61$ or $40$
For $n=40\Rightarrow T_n>0$
For $n=61\Rightarrow T_n<0$
$\therefore n=61$
$\Rightarrow T_n=\frac{100}{5}+(61-1)\left(\frac{-2}{5}\right)=-4$