How many terms of the AP :

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


$a=9, d=8$

Let $S_{n}=636$

$\Rightarrow \frac{n}{2}[2 a+(n-1) d]=636$

$\Rightarrow \frac{n}{2}\{2 \times 9+(n-1)(8)\}=636$

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

$\Rightarrow \frac{n}{2}\{8 n+10\}=636 \Rightarrow n(4 n+5)=636$

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

$\Rightarrow \mathrm{n}=\frac{-5 \pm \sqrt{25+10176}}{8}=\frac{-5 \pm \sqrt{10201}}{8}$

$=\frac{-5 \pm 101}{8}=-\frac{106}{8}$ or $\frac{96}{8}=-\frac{53}{4}$ or 12

We reject $\mathrm{n}=-\frac{53}{4} \Rightarrow \mathrm{n}=12$

Hence, 12 terms makes the sum.

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