Question

If the sum of first $n$ even natural numbers is $420$, find the value of $n$.

Answer 1

We have,
$2+4+6+8+....$ to $n$ tems $=420$

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

here $a=2,d=2,S_n=420$

$\Rightarrow \frac{n}{2}[2\times 2+(n-1)\times 2]=420\Rightarrow n(2+n-1)=420\Rightarrow n(n+1)=420$

$\Rightarrow n^2+n-420=0\Rightarrow n^2+21n-20n-420\Rightarrow n(n+21)-20(n+21)=0$

$\Rightarrow n=20,-21$

$\because$ $x$ can not be a negative

$\Rightarrow n=20$