Question

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

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

Answer 1

Answer: (C)

$S=\frac{100}{5}+\frac{98}{5}+\frac{96}{5}+\frac{94}{5}+\dots n$
$S_n=\frac{n}{2}\left(2\times \frac{100}{5}+(n-1)\left(-\frac{2}{5}\right)\right)=188$
$n(100-n+1)=488\times 5$
$n^2-101n+488\times 5=0$
$n=61,40$
$T_n=a+(n-1)d=\frac{100}{5}-\frac{2}{5}\times 60$
$=20-24=-4$