Discussion
If the variance of \(1, 2, 3, \ldots, n\) is 10, then the value of \(n\) is
Answer: C
\(n\) is 11.
The variance of the first \(n\) natural numbers is \(\frac{n^{2} - 1}{12}\), since the mean of the squares is \(\frac{(n+1)(2n+1)}{6}\) and the square of the mean is \(\frac{(n+1)^{2}}{4}\).
\[ \begin{aligned} \frac{n^{2} - 1}{12} &= 10 \\ n^{2} &= 121, \quad n = 11 \end{aligned} \]
Check: for 1 to 11 the mean is 6 and the squared deviations add to 110, so the variance is \(110 / 11 = 10\). The other values give variances of 2, 6.67, 14 and 18.67.
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