Variance of first n natural numbers {1,2,...,n} = $\frac{n^2-1}{12}$. This is derived from Mean = $\frac{n+1}{2}$ and sum($i^2$) = n(n+1)$\frac{2n+1}{6}$. For an AP {a, a+d, a+2d, ..., a+(n-1)d}: Var = $d^2$$\frac{n^2-1}{12}$. The variance of an AP depends only on the common difference d and the number of terms n, not on the first term a (since shifting doesn't change variance). Example: Var({3,7,11,15,19}) = 4^2$\frac{25-1}{12}$ = 162 = 32. For consecutive even or odd numbers, d=2: Var = 4$\frac{n^2-1}{12}$ = $\frac{n^2-1}{3}$.