Question

Let \(9 =\) \(x_1​<x_2​<…<x_7​\) be 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 :

• A. \(18\left(1+\frac{1}{\sqrt{3}}\right)\)
• B. \(2\left(9+\frac{8}{\sqrt{7}}\right)\)
• C. 34
• D. 25

Hint:

In problems involving sequences and standard deviation, remember to simplify by shifting terms to start at zero and then solve for variance and mean independently.

Answer 1

The Correct Option is C

Given that \(9 = x_1 < x_2 < \cdots < x_7\), the terms of the A.P. are:
\[x_1 = 9, \, x_2 = 9 + d, \, x_3 = 9 + 2d, \, \dots, \, x_7 = 9 + 6d.\]
To simplify, subtract \(9\) from all terms:
\[0, d, 2d, \dots, 6d.\]
The mean is:
\[\bar{x}_{\text{new}} = \frac{0 + d + 2d + \cdots + 6d}{7} = \frac{21d}{7} = 3d.\]
The variance is:
\[\sigma^2 = \frac{1}{7} \left(0^2 + 1^2 + 2^2 + \cdots + 6^2\right) d^2 - \bar{x}_{\text{new}}^2.\]
Using the sum of squares formula:
\[\sum_{k=0}^6 k^2 = \frac{n(n+1)(2n+1)}{6}, \, n = 6.\]
\[\sum_{k=0}^6 k^2 = \frac{6(7)(13)}{6} = 91.\]
Thus:
\[\sigma^2 = \frac{1}{7}(91)d^2 - (3d)^2.\]
\[\sigma^2 = \frac{91}{7}d^2 - 9d^2.\]
\[\sigma^2 = 13d^2 - 9d^2 = 4d^2.\]
The standard deviation is given as \(4\):
\[\sqrt{4d^2} = 4 \implies d^2 = 4 \implies d = 2.\]
Now, calculate \(\bar{x} + x_6\):
\[\bar{x} = 3d + 9 = 3(2) + 9 = 15.\]
\[x_6 = 9 + 5d = 9 + 5(2) = 19.\]
\[\bar{x} + x_6 = 15 + 19 = 34.\]
Conclusion: \(\bar{x} + x_6 = 34\)