Question

In an arithmetic progression $2,7,12,17,......$ summation of how many terms will be $990$?

Answer 1

For the given A.P $2,7,12,17,....$ we have $a=2,d=7-2=5$ and $S_n=990$
$S_n=\frac{1}{2}n[2a+(n-1\left)d\right]$
$990=\frac{1}{2}n[4+(n-1\left)5\right]$
$1980=n(5n-1)$
$1980=5n^2-n$
$5n^2-n-1980=0$
$(5n+99)(n-20)=0$
$⟹n=\frac{-99}{5}$ or $n=20$
Since $n\in N,n=\frac{-99}{5}$ is not possible
$\therefore$ $n=20$
Thus $20$ terms of the given A.P add upto $990$