Question

How many terms of the AP 9, 17, 25 … must be taken to give a sum of 636?

Answer 1

Let there be n terms in this A.P.
For this A.P, a = 9
$d=a_2-a_1=17-9=8As{S}_n=\frac{n}{2}[2a+(n-1\left)d\right]636=\frac{n}{2}[2\times a+(n-1)\times 8]636=\frac{n}{2}[18+(n-1)\times 8]636=n[9+4n-4]636=n(4n+5)4n^2+5n-636=04n^2+53n-48n-636=0n(4n+53)-12(4n+53)=0(4n+53)(n-12)=0Either4n+53=0orn-12=0n=\left(\frac{-53}{4}\right)orn=12$
As the number of terms can neither be negative nor fractional, n = 12.