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 x , then x + x 6 is equal to [2023]

Question

Let $9 = x_{1} < x_{2} < \dots < x_{7}$ be in an A.P. with common difference $d$ . If the standard deviation of $x_{1}, x_{2}, \dots, x_{7}$ is $4$ and the mean is $\bar{x}$ , then $\bar{x} + x_{6}$ is equal to [2023]

Smart Achievers · Accepted Answer

(3)

$Mean = \bar{x} = \frac{\sum_{i = 1}^{7} x_{i}}{7} = \frac{\frac{7}{2} [2 a + 6 d]}{7} = a + 3 d = x_{4}$

$Variance = \frac{\sum_{i = 1}^{7} {(x_{i} - \bar{x})}^{2}}{7} = {(4)}^{2} \Rightarrow \frac{\sum_{i = 1}^{7} {(x_{i} - x_{4})}^{2}}{7} = 16$

$\Rightarrow \frac{{(3 d)}^{2} + {(2 d)}^{2} + d^{2} + 0 + d^{2} + {(2 d)}^{2} + {(3 d)}^{2}}{7} = 16$

$\Rightarrow \frac{28 d^{2}}{7} = 16 \Rightarrow d = 2 \Rightarrow \bar{x} = 9 + 3 (2) = 15$

$x_{6} = a + 5 d = 9 + 5 (2) = 19$ $∴$ $\bar{x} + x_{6} = 15 + 19 = 34$

Solution

Q.
Let $9 = x_{1} < x_{2} < \dots < x_{7}$ be in an A.P. with common difference $d$ . If the standard deviation of $x_{1}, x_{2}, \dots, x_{7}$ is $4$ and the mean is $\bar{x}$ , then $\bar{x} + x_{6}$ is equal to [2023]

1 $18 (1 + \frac{1}{\sqrt{3}})$

2 $2 (9 + \frac{8}{\sqrt{7}})$

3 $34$

4 $25$

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Solution

Q. Let 9=x1<x2<⋯<x7 be in an A.P. with common difference d. If the standard deviation of x1,x2,…,x7 is 4 and the mean is x¯, then x¯+x6 is equal to [2023]

1 18(1+13)

2 2(9+87)

3 34

4 25