Question:
If the variance of the first $n$ natural numbers is 10 and the variance of the first $m$ even natural numbers is 16 , then $m+n$ is equal to__________.
Solution:
$\operatorname{Var}(1,2, \ldots . ., n)=10$
$\Rightarrow \frac{1^{2}+2^{2}+\ldots . .+n^{2}}{n}-\left(\frac{1+2+\ldots . .+n}{n}\right)^{2}=10$
$\Rightarrow \frac{(n+1)(2 n+1)}{6}-\left(\frac{n+1}{2}\right)^{2}=10$
$\Rightarrow n^{2}-1=120 \Rightarrow n=11$
$\operatorname{Var}(2,4,6, \ldots . ., 2 m)=16 \Rightarrow \operatorname{Var}(1,2, \ldots . ., m)=4$
$\Rightarrow m^{2}-1=48 \Rightarrow m=7$
$\Rightarrow m+n=18$