Question

The sum of n terms of an A.P. is $136$, its common difference is $4$ and last term is $31$. Find the value of n.

Answer 1

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

$a+(n-1)d=31\Rightarrow a=31-4(n-1)$

$\therefore 136=\frac{n}{2}\left\{2\right[31-4(n-1)]+(n-1\left)d\right\}$

$n[70-8n+4n-4]=272$

$n(66-4n)=272$

$4n^2-66n+272=0$

$2n^2-33n+136=0$

$2n(n-8)-17(n-8)=0$

$(2n-17)(n-8)=0$

$\therefore n=8$