Question

If variance of first $n$ natural number is $10$ and variance of first $m$ even natural number is $16$, then the value of $m+n$ is

Answer 1

$\text{Variance}=\frac{\sum _{i=1}^n{x}_i^2}{n}-(\overline{x}{)}^2$
$10=\frac{n(n+1)(2n+1)}{6n}-{\left[\frac{n(n+1)}{2n}\right]}^2$
$=\frac{(n+1)(2n+1)}{6}-\frac{(n+1{)}^2}{4}$
$=\frac{(n+1)}{24}\left[4\right(2n+1)-6(n+1\left)\right]$
$\Rightarrow (n+1)(n-1)=\frac{240}{2}$
$\Rightarrow n^2-1=120$
$\Rightarrow n=11$

Similarly, first $m$ even numbers are $2,4,6,\cdots 2m,$ we have
$16=\frac{4(1^2+2^2+3^2+4^2+\cdots +m^2)}{m}-{\left[\frac{2(1+2+3+4+\cdots \cdots +m)}{m}\right]}^2$
$=\frac{4m(m+1)(2m+1)}{6m}-{\left[\frac{2m(m+1)}{2m}\right]}^2$
$\Rightarrow (m+1)\left[\frac{2}{3}\right(2m+1)-(m+1\left)\right]=16$
$\Rightarrow (m+1)(m-1)=48$ $\Rightarrow m^2-1=48$
$\Rightarrow m=7$
$\therefore m+n=7+11=18$