Question

How many terms of the AP 9, 17, 25, ... must be taken so that their sum is 636?

Answer 1

A.P = 9,17,25
a=9
d=17-9 =8

Sum of n terms, $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

$636=\frac{n}{2}\left[2\right(9)+(n-1\left)8\right]$

$636=\frac{n}{2}[18+8n-8]$

$636=\frac{n}{2}[10+8n]$

$1272=10n+8n^2$

$8n^2+10n-1272=0$

$2[4n^2+5n-636]=0$

$4n^2+5n-636=0$

$4n^2+53n-48n-636=0$

$n(4n+53)-12(4n+53)=0$

$(n-12)(4n+53)=0$

$n-12=0$

$n=12$

12 terms must be given to sum of 636