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 are given $$9 = x_1 < x_2 < \ldots < x_7$$ in an AP with common difference $$d$$, standard deviation = 4, and mean = $$\bar{x}$$, and we need to find $$\bar{x} + x_6$$.

First, express the terms of the arithmetic progression by noting that for $$k = 1,2,\ldots,7$$ we have $$x_k = 9 + (k-1)d$$.

Next, recall that for an AP with an odd number of terms, the mean equals the middle term, so $$\bar{x} = x_4 = 9 + 3d$$.

To find the variance, observe that the terms differ by multiples of $$d$$, so the variance of the set $$\{x_1,x_2,\ldots,x_7\}$$ is the same as the variance of $$\{0,d,2d,3d,4d,5d,6d\}$$, namely $$\sigma^2 = d^2 \cdot \text{Var}(\{0,1,2,3,4,5,6\})$$. Since $$\text{Var}(\{0,1,2,3,4,5,6\}) = \frac{0^2 + 1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2}{7} - \left(\frac{0+1+2+3+4+5+6}{7}\right)^2 = \frac{91}{7} - \left(\frac{21}{7}\right)^2 = 13 - 9 = 4\,,$$ it follows that $$\sigma^2 = 4d^2$$.

Given that $$\sigma = 4$$ implies $$\sigma^2 = 16$$, we set $$4d^2 = 16$$ which leads to $$d^2 = 4$$ and hence $$d = 2$$, taking the positive value since the sequence is increasing.

Now substituting back gives $$\bar{x} = 9 + 3(2) = 15$$ and $$x_6 = 9 + 5(2) = 19$$.

Therefore, $$\bar{x} + x_6 = 15 + 19 = 34$$.

The correct answer is Option B: 34.