Let 9 < x₁ < x₂ < ldots < xg be in an A.P. with common difference d . If s

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13.Statistics

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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 $\overline{ x }$, then $\overline{ x }+ x _6$ is equal to:

A

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

B

$34$

C

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

D

$25$

(JEE MAIN-2023)

Solution

$9=x_1 < x_2 < \ldots \ldots < x_7$

$9,9+d, 9+2 d, \ldots \ldots .9+6 d$

$0, d, 2 d, \ldots \ldots \cdot 6$

$\bar{x}_{\text {new }}=\frac{21 d }{7}=3 d$

$16=\frac{1}{7}\left(0^2+1^2+\ldots \ldots+6^2\right) d^2-9 d^2$

$=\frac{1}{7}\left(\frac{6 \times 7 \times 13}{6}\right) d ^2-9 d ^2$

$16=4 d^2$

$d^2=4$

$d=2$

$\bar{x}+x_6=6+9+10+9$

Standard 11

Mathematics

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