How many terms of the A.P. $9, 17, 25,…$ must be taken so that their sum is 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,

$\mathrm{S}_{n}=\frac{n}{2}[2 a+(n-1) d]$
$\Rightarrow 636=\frac{n}{2}[2 \times 9+(n-1) \times 8]$
$\Rightarrow 1272=n[18+8 n-8]$
$\Rightarrow 1272=n(10+8 n)$
$\Rightarrow 1272=10 n+8 n^{2}$

$\Rightarrow 8 n^{2}+10 n-1272=0$

$\Rightarrow 2(4 n^{2}+5 n-636)=0$
$\Rightarrow 4 n^{2}+5 n-636=0$
$\Rightarrow 4 n^{2}+53 n-48 n-636=0$
$\Rightarrow n(4 n+53)-12(4 n+53)=0$
$\Rightarrow(4 n+53)(n-12)=0$
\( 4 n+53=0 \) or \( n-12=0 \)

\( n=\frac{-53}{4} \) which is not possible or \( n=12 \)

\( \therefore \) \( n=12 \)

The number of terms to be taken is 12.  

Tutorialspoint

Simply Easy Learning

Updated on: 10-Oct-2022

58 Views

Kickstart Your Career

Get certified by completing the course

Get Started