This problem combines concepts from Arithmetic Progressions (A.P.) and basic statistics, specifically mean and standard deviation.

1. Key Concepts and Formulas

• Arithmetic Progression (A.P.): A sequence where the difference between consecutive terms is constant, known as the common difference ($d$). The $n$-th term is $x_n = a + (n-1)d$, where $a$ is the first term.
• Mean ($\bar{x}$): The average of a set of $n$ numbers, $\bar{x} = \frac{\sum_{i=1}^n x_i}{n}$. For an A.P. with an odd number of terms, the mean is the middle term.
• Variance ($\sigma^2$) and Standard Deviation ($\sigma$): Variance measures the spread of data and is defined as $\sigma^2 = \frac{\sum_{i=1}^n (x_i - \bar{x})^2}{n}$. Standard deviation is the square root of the variance, $\sigma = \sqrt{\sigma^2}$. For an A.P. with $n$ terms and common difference $d$, the variance can also be calculated using the shortcut formula $\sigma^2 = \frac{d^2(n^2-1)}{12}$.

2. Step-by-Step Solution

We are given:

• First term: $x_1 = 9$. So, $a=9$.
• Number of terms: $n=7$.
• Standard deviation: $\sigma = 4$.
• Terms are increasing: $x_1 < x_2 < \ldots < x_7$, which implies $d > 0$. Our goal is to find the value of $\bar{x} + x_6$.

Step 1: Expressing the Terms and Mean of the A.P.

The terms of the A.P. are: $x_1 = a = 9$ $x_2 = a+d = 9+d$ ... $x_7 = a+6d = 9+6d$

For an A.P. with an odd number of terms, the mean ($\bar{x}$) is the middle term. Here, $n=7$, so the middle term is $x_4$. $\bar{x} = x_4 = a+3d = 9+3d$

Step 2: Using the Standard Deviation to Find the Common Difference ($d$)

We are given $\sigma = 4$. Therefore, the variance $\sigma^2 = 4^2 = 16$. We can use the shortcut formula for the variance of an A.P.: $\sigma^2 = \frac{d^2(n^2-1)}{12}$ Substitute $n=7$ and $\sigma^2=16$: $16 = \frac{d^2(7^2-1)}{12}$ $16 = \frac{d^2(49-1)}{12}$ $16 = \frac{d^2(48)}{12}$ $16 = 4d^2$ Divide by 4: $d^2 = \frac{16}{4} = 4$ Take the square root: $d = \pm \sqrt{4} = \pm 2$ Since $x_1 < x_2 < \ldots < x_7$, the common difference $d$ must be positive. Therefore, $d=2$.

Step 3: Calculating $\bar{x}$ and $x_6$

Now that we have $a=9$ and $d=2$, we can calculate the mean and the sixth term:

• Mean ($\bar{x}$): $\bar{x} = 9+3d = 9+3(2) = 9+6 = 15$
• Sixth term ($x_6$): $x_6 = a+5d = 9+5(2) = 9+10 = 19$

Step 4: Final Calculation: $\bar{x} + x_6$

Finally, we sum the calculated values: $\bar{x} + x_6 = 15 + 19 = 34$

3. Common Mistakes & Tips

• Sign of common difference: Always check the condition like $x_1 < x_2 < \ldots < x_7$ to determine if $d$ is positive or negative.
• Mean of an A.P.: For an A.P. with an odd number of terms, the mean is simply the middle term. This is a powerful shortcut.
• Variance of an A.P. Shortcut: Remember the formula $\sigma^2 = \frac{d^2(n^2-1)}{12}$ for an A.P. This saves significant time compared to calculating $\sum (x_i - \bar{x})^2$ term by term.

4. Summary

We started by identifying the given information about the arithmetic progression and its standard deviation. We used the property that the mean of an A.P. with an odd number of terms is its middle term. The standard deviation formula for an A.P. allowed us to efficiently calculate the common difference $d$. With the first term and common difference determined, we then calculated the mean and the sixth term, and finally their sum. The calculation yields a sum of 34.

5. Final Answer

The final answer is $\boxed{34}$, which corresponds to option (D).