If the variance of the first $n$ natural numbers is 10 and the variance of the first $m$ even natural numbers is 16, then the value of $m + n$ is equal to
Step-by-Step Solution
Key Concept: Covariance of arithmetic progressions can be computed using the standard formula involving the sum of squares.
Given $\text{cov}(1,2,\ldots,n) = 10$ with the covariance formula $\text{cov} = \frac{1^2 + 2^2 + \cdots + n^2}{n} - \left(\frac{n+1}{2}\right)^2 = 10$. Using $\sum k^2 = \frac{n(n+1)(2n+1)}{6}$, we get $\frac{(n+1)(2n+1)}{6} - \frac{(n+1)^2}{4} = 10$. Simplifying yields $n^2 - 1 = 120$, so $n = 11$. For $\text{cov}(4, 8, \ldots, 2m) = 4$, we apply the same formula to get $m^2 - 1 = 48$, so $m = 7$, giving $m + n = 18$.
Correct Answer: 18