Question

Find the number of terms of the following AP. $9,17,25,....$ having sum $636$.

Answer 1

Step 1. Find the quadratic equation by using the formula of sum of first $n^{th}$ terms of an AP:

The given AP is $9,17,25,....$

Here, first term, $a=9$ and common difference, $d=17-9=8$

Let $n$ terms of the AP must be taken, then

$S_n=636$

We know that the sum of first $n^{th}$ terms of an AP is $S_n=\frac{n}{2}\left[2a+(n-1)d\right]$.

$\begin{array}{cc}\Rightarrow & \frac{n}{2}\left[2\times 9+(n-1)\times 8\right]=636\\ \Rightarrow & n\left[9+4(n-1)\right]=636\\ \Rightarrow & n\left[9+4n-4)\right]=636\\ \Rightarrow & n\left[4n+5\right]=636\\ \Rightarrow & 4n^2+5n=636\\ \Rightarrow & 4n^2+5n-636=0\end{array}$

Step 2. Solve the quadratic equation by factorization method:

$$\begin{gathered} \begin{array}{cc}\Rightarrow & 4n^2+5n-636=0\\ \Rightarrow & 4n^2+53n-48n-636=0\\ \Rightarrow & n(4n+53)-12(4n+53)=0\\ \Rightarrow & (4n+53)(n-12)=0\end{array} \\ \begin{array}{cc}\Rightarrow & (4n+53)=0and(n-12)=0\\ \Rightarrow & n=-\frac{53}{4}=0andn=12\end{array} \end{gathered}$$

Step 3. Find the possible number of terms:

$n=-\frac{53}{4}$is neglected as $n$ being the number of terms, is a natural number.

$\therefore n=12$

Hence, the sum of first $12$ terms of the AP is $636$.