Question

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

Answer 1

$a=9,d=8$

$S_n=636=\frac{n}{2}(18+8(n-1\left)\right)$

$\Rightarrow 1272=10n+8n^2$

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

$Root=\frac{b\pm \sqrt{b^2-4ac}}{2a}$

$=\frac{5\pm \sqrt{25+10176}}{8}=\frac{5\pm \sqrt{10201}}{8}=\frac{5\pm 101}{8}$

$\alpha =\frac{106}{8}=\frac{53}{4}\beta =\frac{-96}{8}=-12$

As twelve is the integer so number of terms will be 12.