Question

If the sum of $n$ successive odd natural numbers starting from $3$ is $48$, find the value of $n$.

Answer 1

We have
$3+5+7+9+.....$ to $n$ terms $=48$

We know that

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

here $a=3,d=2,S_n=48$

$\Rightarrow \frac{n}{2}[2\times 3+(n-1)\times 2]=48$

$\Rightarrow n(3+n-1)=48\Rightarrow n^2+2n-48=0\Rightarrow n^2+8n-6n-48=0$

$\Rightarrow n(n+8)-6(n+8)\Rightarrow (n+8)(n-6)=0\Rightarrow n=-8orn=6$

$\because$ $n$ can not be a negative here

$\therefore$ $n=6$