JEE Main · 2020 · 03 Sep 2020 · Shift 2 · Sequences and Series

The terms $20,19\frac{3}{5},19\frac{1}{5},18\frac{4}{5},\dots$ form an arithmetic progression with first term $20$ and common difference $-\frac{2}{5}$. Setting the sum to $488$, $\frac{n}{2}\left[40+(n-1)\left(-\frac{2}{5}\right)\right]=488$ tidies to $n(202-2n)=4880$, i.e. $n^2-101n+2440=0$, with roots $n=40$ and $n=61$. The $n$th term is $20+(n-1)\left(-\frac{2}{5}\right)$; at $n=40$ it is still positive, while at $n=61$ it is $20-24=-4$, which is negative. Since the $n$th term must be negative, $n=61$ and that term is $-4$.