JEE Main · 2025 · 24 Jan 2025 · Shift 1 · Sequences and Series

The general term of the series is $\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$, so the sum telescopes:




For the A.P. with first term $-p$ and common difference $p$, the sum of the first six terms is

Setting $9p=45$ gives $p=5$.

The general term is $a_n=-p+(n-1)p=p(n-2)=5(n-2)$, so $a_{20}=5(18)=90$ and $a_{15}=5(13)=65$.