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