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

AcademicMathematicsNCERTClass 10


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. 


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,


The number of terms to be taken is 12. 

Updated on 10-Oct-2022 13:20:30