Variance of first n natural numbers {1,2,...,n} = $\frac{n^2-1}{12}$. Derivation: Mean = $\frac{n+1}{2}$, sum of squares = n(n+1)$\frac{2n+1}{6}$, Var = n(n+1)$\frac{2n+1}{(6n)}$ - [$\frac{n+1}{2}$]^2 = $\frac{n^2-1}{12}$. For an AP with first term a, common difference d, n terms: Var = $d^2$*$\frac{n^2-1}{12}$. The variance depends only on d and n, not on a (since shifting by 'a' doesn't change variance).