Question

In an A.P. $a=-2,d=4$ and $S_n=160$, find $n$.

Answer 1

In AP , given that $a=-2$ and $d=4$

Given that the sum of first $n$ terms in the series is $S_n=160$

We know that the sum of first $n$ terms in the series is $S_n=\frac{n}{2}(2a+(n-1\left)d\right)$

$\Rightarrow \frac{n}{2}(2a+(n-1\left)d\right)=160$

$\Rightarrow \frac{n}{2}(-4+4(n-1\left)\right)=160$

$\Rightarrow n(n-2)=80$

$\Rightarrow n=10,-8$

Since $n$ is positive integer , the value of $n=10$