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13.Statistics
hard
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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