Question

How many terms of the arithmetic progression $9,17,25$… must be taken to give a sum of $636$?

Answer 1

Step 1: Find the common difference of the arithmetic progression

$\begin{array}{rcl}d& =& a_2-a_1\\ & =& 17-9\\ & \Rightarrow & d=8\end{array}$

Step 2 : Find the number of terms in the arithmetic progression

The sum of $n$terms of an arithmetic progression is given by

$S_n=\frac{n}{2}\left[2a+(n-1)d\right]$

Substituting the values of $S_n,a,d$ in the standard result we get

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

$\Rightarrow 636=n(9+4n-4)$

$\Rightarrow 636=4n^2+5n$

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

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

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

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

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

As $n$cannot be a negative rational number, $n=12$

Hence, the number of terms of the given arithmetic progression is $12$