If the variance of the first n natural number | vedclass

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Question

(B) The variance of the first $n$ natural numbers is given by the formula $\frac{n^{2}-1}{12}$.Given $\frac{n^{2}-1}{12} = 10$,we have $n^{2}-1 = 120$

Vedclass · Accepted Answer

$18$

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(B) The variance of the first $n$ natural numbers is given by the formula $\frac{n^{2}-1}{12}$.Given $\frac{n^{2}-1}{12} = 10$,we have $n^{2}-1 = 120$,so $n^{2} = 121$,which gives $n = 11$.The variance of the first $m$ even natural numbers $(2, 4, 6, ..., 2m)$ is $4$ times the variance of the first $m$ natural numbers.Thus,the variance is $4 	imes \frac{m^{2}-1}{12} = \frac{m^{2}-1}{3}$.Given $\frac{m^{2}-1}{3} = 16$,we have $m^{2}-1 = 48$,so $m^{2} = 49$,which gives $m = 7$.Therefore,$m + n = 7 + 11 = 18$.

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

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