How many terms of AP: $9, 17, 25, …$ must be taken to give a sum of 636?
Given:
Given A.P. is $9, 17, 25,…$
To do:
We have to find the number of terms that must be taken so that their sum is 636.
Solution:
Let the number of terms of the A.P. be $n$.
First term $(a)=9$
Common difference $(d)=17-9=8$
Sum of the $n$ terms $=636$
We know that,
$\mathrm{S}_{n}=\frac{n}{2}[2 a+(n-1) d]$
$\Rightarrow 636=\frac{n}{2}[2 \times 9+(n-1) \times 8]$
$1272=n[18+8 n-8]$
$1272=n(10+8 n)$
$1272=10 n+8 n^{2}$
$8 n^{2}+10 n-1272=0$
$2(4 n^{2}+5 n-636)=0$
$4 n^{2}+5 n-636=0$
$4 n^{2}+53 n-48 n-636=0$
$n(4 n+53)-12(4 n+53)=0$
$(4 n+53)(n-12)=0$
$4n+53=0$ or $n-12=0$
$n=\frac{-53}{4}$ which is not possible or $n=12$
This implies,
$n=12$
The number of terms to be taken is 12.  
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