Question

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

Answer 1

Let there be $n$ terms of this A.P.
For this A.P., $a=9$

$d=17-9=8$

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

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

$636=n[9+4n-4]$

$636=n(4n+5)$

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

$4n^2+53n-48n-636=0$

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

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

Either $4n+53=0$ or $n-12=0$

$n=\frac{53}{4}$ or $n=12$

$n$ cannot be $\frac{53}{4}$.(As the number of terms can neither be negative nor fractional.)

Therefore, $n=12$ only.