If the variance of first $n$ even natural numbers is $133$, then the value of $n$ is equal to
Step-by-Step Solution
Key Concept: For an arithmetic sequence, variance can be expressed in terms of the number of terms; equate to the given variance to solve for $n$.
Given $\text{Var}(2, 4, 6, \ldots, 2n) = 133$. The general term is $2i$ for $i = 1, 2, \ldots, n$. Mean: $\bar{x} = \frac{\sum_{i=1}^{n} 2i}{n} = \frac{2 \cdot \frac{n(n+1)}{2}}{n} = n + 1$. Variance: $\sigma^2 = \frac{\sum_{i=1}^{n} (2i)^2}{n} - (n+1)^2 = \frac{4\sum_{i=1}^{n} i^2}{n} - (n+1)^2 = \frac{4 \cdot \frac{n(n+1)(2n+1)}{6}}{n} - (n+1)^2 = \frac{2(n+1)(2n+1)}{3} - (n+1)^2 = (n+1)\left(\frac{2(2n+1) - 3(n+1)}{3}\right) = \frac{(n+1)(n-1)}{3}$. Setting this equal to 133: $\frac{(n+1)(n-1)}{3} = 133$, so $(n+1)(n-1) = 399 = n^2 - 1$, giving $n^2 = 400$, thus $n = 20$.
Correct Answer: 4