If the variance of the first $n$ natural numbers is 10 and the variance of the first $m$ even natura

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

Professor PC Thomas Classes · Answer

(18): Given, variance of first $n$ natural numbers $=10$\ $\Rightarrow \frac{1^{2}+2^{2}+\ldots+n^{2}}{n}-\left(\frac{1+2+\ldots+n}{n}ight)^{2}=10$\ $\Rightarrow \frac{(n+1)(2 n+1)}{6}-\left(\frac{n+1}{2}ight)^{2}=10$\ $\Rightarrow \frac{n^{2}-1}{12}=10$\ $\Rightarrow \quad n^{2}-1=120 \Rightarrow n=11$\ Now, var $(2,4,6, \ldots \ldots, 2 m)=16$\ [Given]\ $\Rightarrow \quad \operatorname{Var}(1,2, \ldots \ldots, m)=\frac{16}{2^{2}}=4$\ $\Rightarrow \quad m^{2}-1=48 \Rightarrow m=7$\ $	herefore \quad m+n=18$\