Let `9 = x_1 lt x_2 lt ……lt x_7` in an A.P. with common difference d. If the standard deviation of `x_1, x_2 …… x_7` is 4 and the mean is `bar x` , then `bar x + x_6` is equal to

Options

A. `18(1+(1)/(sqrt3))`
B. `2(9+(8)/(sqrt7))`
C. `34`
D. `25`

Answer

To solve the problem step by step, we will follow the given information and apply the necessary formulas.

### Step 1: Understand the Arithmetic Progression (A.P.)
We are given that \( x_1, x_2, \ldots, x_7 \) are in an A.P. with a common difference \( d \). The terms can be expressed as:
- \( x_1 = a \)
- \( x_2 = a + d \)
- \( x_3 = a + 2d \)
- \( x_4 = a + 3d \)
- \( x_5 = a + 4d \)
- \( x_6 = a + 5d \)
- \( x_7 = a + 6d \)

### Step 2: Calculate the Mean
The mean \( \bar{x} \) of these 7 terms is given by:
\[ \bar{x} = \frac{x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7}{7} \]
Substituting the expressions for \( x_i \):
\[ \bar{x} = \frac{a + (a + d) + (a + 2d) + (a + 3d) + (a + 4d) + (a + 5d) + (a + 6d)}{7} \]
\[ = \frac{7a + (0 + 1 + 2 + 3 + 4 + 5 + 6)d}{7} \]
\[ = \frac{7a + 21d}{7} = a + 3d \]

### Step 3: Calculate the Standard Deviation
The standard deviation \( \sigma \) is given as 4. The variance \( \sigma^2 \) is therefore \( 4^2 = 16 \).

The formula for variance is:
\[ \sigma^2 = \frac{1}{n} \sum_{i=1}^{n} (x_i - \bar{x})^2 \]
For our case:
\[ 16 = \frac{1}{7} \left( (x_1 - \bar{x})^2 + (x_2 - \bar{x})^2 + (x_3 - \bar{x})^2 + (x_4 - \bar{x})^2 + (x_5 - \bar{x})^2 + (x_6 - \bar{x})^2 + (x_7 - \bar{x})^2 \right) \]
This implies:
\[ 112 = (x_1 - \bar{x})^2 + (x_2 - \bar{x})^2 + (x_3 - \bar{x})^2 + (x_4 - \bar{x})^2 + (x_5 - \bar{x})^2 + (x_6 - \bar{x})^2 + (x_7 - \bar{x})^2 \]

### Step 4: Substitute \( x_i \) and Simplify
Substituting \( \bar{x} = a + 3d \):
- \( x_1 - \bar{x} = a - (a + 3d) = -3d \)
- \( x_2 - \bar{x} = -2d \)
- \( x_3 - \bar{x} = -d \)
- \( x_4 - \bar{x} = 0 \)
- \( x_5 - \bar{x} = d \)
- \( x_6 - \bar{x} = 2d \)
- \( x_7 - \bar{x} = 3d \)

Calculating the squares:
\[ (-3d)^2 + (-2d)^2 + (-d)^2 + 0^2 + d^2 + (2d)^2 + (3d)^2 = 9d^2 + 4d^2 + d^2 + 0 + d^2 + 4d^2 + 9d^2 = 28d^2 \]
Setting this equal to 112:
\[ 28d^2 = 112 \implies d^2 = 4 \implies d = 2 \]

### Step 5: Find \( a \) and Calculate \( \bar{x} \)
Now substituting \( d \) back into the mean:
\[ \bar{x} = a + 3d = a + 3(2) = a + 6 \]

### Step 6: Calculate \( x_6 \)
Now, calculate \( x_6 \):
\[ x_6 = a + 5d = a + 5(2) = a + 10 \]

### Step 7: Find \( \bar{x} + x_6 \)
Now we can find \( \bar{x} + x_6 \):
\[ \bar{x} + x_6 = (a + 6) + (a + 10) = 2a + 16 \]

### Step 8: Determine \( a \)
Since \( x_1 < x_2 < \ldots < x_7 \) and \( x_1 = a \), we know \( a \) must be less than \( x_6 \). We can assume \( a = 9 \) (as given in the problem). Thus:
\[ \bar{x} + x_6 = 2(9) + 16 = 18 + 16 = 34 \]

### Final Answer
Thus, \( \bar{x} + x_6 = 34 \).