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

The given AP is 9, 17, 25, ... .

Here, a = 9 and d = 17 − 9 = 8

Let the required number of terms be n. Then,

$S_{n}=636$

$\Rightarrow \frac{n}{2}[2 \times 9+(n-1) \times 8]=636 \quad\left\{S_{n}=\frac{n}{2}[2 a+(n-1) d]\right\}$

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

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

$\Rightarrow n(5+4 n)=636$

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

$\Rightarrow 4 n^{2}-48 n+53 n-636=0$

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

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

$\Rightarrow n-12=0$ or $4 n+53=0$

$\Rightarrow n=12$ or $n=-\frac{53}{4}$

∴ n = 12                    (Number of terms cannot negative)

Hence, the required number of terms is 12.