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.
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