Let the six numbers $$a_1, a_2, \ldots, a_6$$ be in A.P. and $$a_1 + a_3 = 10$$. If the mean of these six numbers is $$\frac{19}{2}$$ and their variance is $$\sigma^2$$, then $$8\sigma^2$$ is equal to

Let the six numbers be $$a, a+d, a+2d, a+3d, a+4d, a+5d$$.

$$a + (a + 2d) = 10 \implies 2a + 2d = 10 \implies a + d = 5 \implies a = 5 - d$$

$$\mu = \frac{a + (a+d) + (a+2d) + (a+3d) + (a+4d) + (a+5d)}{6} = \frac{6a + 15d}{6} = a + 2.5d$$

$$\frac{19}{2} = (5 - d) + 2.5d$$

$$9.5 = 5 + 1.5d \implies 4.5 = 1.5d \implies d = 3$$

$$a = 5 - 3 = 2$$

The six numbers are $$2, 5, 8, 11, 14, 17$$.

Using the formula for the variance of $$n$$ terms in an A.P.:

$$\sigma^2 = \frac{(n^2 - 1)d^2}{12}$$

$$\sigma^2 = \frac{(6^2 - 1) \times 3^2}{12} = \frac{35 \times 9}{12} = \frac{35 \times 3}{4} = \frac{105}{4}$$

$$8\sigma^2 = 8 \times \left( \frac{105}{4} \right) = 2 \times 105 = 210$$