The variance of the numbers 8, 21, 34, 47,…, 320 is

We need to find the variance of the arithmetic progression 8, 21, 34, 47, ..., 320.

The first term is a = 8 and the common difference is d = 13. Since the last term is 320, we set $$a + (n-1)d = 320$$ which leads to $$8 + (n-1)(13) = 320$$ and hence $$(n-1) = \frac{312}{13} = 24$$ so that $$n = 25$$.

For an AP with n terms and common difference d, the variance is given by $$\sigma^2 = \frac{d^2(n^2 - 1)}{12}$$. Substituting d = 13 and n = 25 yields

$$\sigma^2 = \frac{13^2(25^2 - 1)}{12} = \frac{169 \times 624}{12} = 169 \times 52 = 8788$$.

The answer is 8788.