Question

How many terms of the ap 3 5 7 9 must be added to get the sum 120?

Answer 1

Given AP is 3, 5, 7, 9,...

$a=3\quad;\quad d=5-3=2$

We have,

$S_n=\dfrac{n}{2}[2a+(n-1)d]\\\\\\S_n=\dfrac{n}{2}[2\times3+(n-1)2]\\\\\\S_n=\dfrac{n}{2}[6+2n-2]\\\\\\S_n=\dfrac{n}{2}[2n+4]\\\\\\S_n=n(n+2)\\\\\\S_n=n^2+2n$

Hence we got the sum of first 'n' terms.

Let the sum be 120. So we have to find 'n'.

$\begin{aligned}&n^2+2n=120\\\\\implies\ \ &n^2+2n+1=120+1\\\\\implies\ \ &(n+1)^2=121\\\\\implies\ \ &n+1=\pm11\\\\\implies\ \ &n=10\quad\ \ \text{OR}\ \ \quad n=-12\end{aligned}$

But  $n\geq 1.$  So  $n\neq-12.$

Hence  $\mathbf{n=10}$