Let $$9 = x_1 < x_2 < \ldots < x_7$$ be in an A.P. with common difference $$d$$. If the standard deviation of $$x_1, x_2, \ldots, x_7$$ is 4 and the mean is $$\bar{x}$$, then $$\bar{x} + x_6$$ is equal to :

We have 7 terms in A.P.: $$x_1 = 9, x_2 = 9+d, \ldots, x_7 = 9+6d$$. The mean of these terms is $$\bar{x} = \frac{x_1 + x_2 + \cdots + x_7}{7} = \frac{7 \cdot 9 + d(0+1+2+3+4+5+6)}{7} = 9 + 3d$$. The deviations from the mean are $$-3d, -2d, -d, 0, d, 2d, 3d$$, so the variance is $$\sigma^2 = \frac{1}{7}\left(9d^2 + 4d^2 + d^2 + 0 + d^2 + 4d^2 + 9d^2\right) = \frac{28d^2}{7} = 4d^2$$. Given $$\sigma = 4$$, we have $$4d^2 = 16 \implies d^2 = 4 \implies d = 2 \text{ (since terms are increasing)}$$. Then $$\bar{x} = 9 + 3(2) = 15$$ and $$x_6 = 9 + 5(2) = 19$$, so $$\bar{x} + x_6 = 15 + 19 = 34$$.

The correct answer is Option B: 34.