Question

How many terms of the A.P. $9,17,25,...$ must be taken so that their sum is $636$?

Answer 1

Calculate the number of terms of the given A.P.

Given, $9,17,25,...$ are in A.P.

First term $\left(a\right)=9$

Common difference $\left(d\right)=17-9=8$

Sum of $n$ terms $\left(S_n\right)=636$

We know that, Sum of $n$ terms, $S_n=\frac{n}{2}\left[2a+(n-1)d\right]$

Substituting the values in $S_n$, we get

$$\begin{gathered} \Rightarrow 636=\frac{n}{2}\left[2\left(9\right)+(n-1)\left(8\right)\right] \\ \Rightarrow 636=\frac{n}{2}\left[18+(8n-8)\right] \\ \Rightarrow 636\times 2=n\times [10+8n] \\ \Rightarrow 1272=10n+8n^2 \\ \Rightarrow 8n^2+10n–1272=0 \\ \Rightarrow 4n^2+5n–636=0 \\ \Rightarrow 4n^2–48n+53n–636=0 \\ \Rightarrow 4n(n–12)+53(n–12)=0 \\ \Rightarrow (4n+53)(n–12)=0 \end{gathered}$$

So, $n=\frac{-53}{4}$ or $n=12$

Since, the number of terms cannot be negative.

$\therefore n=12$

Hence, the number of terms are $12$.