If variance of terms in an increasing A.P. , b _{1}, b _{2}, b _{3}, ldots b _{11} is 90, then commo

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8. Sequences and Series

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If the variance of the terms in an increasing $A.P.$, $b _{1}, b _{2}, b _{3}, \ldots b _{11}$ is $90,$ then the common difference of this $A.P.$ is

A

$3$

B

$9$

C

$-9$

D

$-3$

(JEE MAIN-2020)

Solution

Let a be the first term and $d$ be the common

difference of the given A.P. Where $d>0$

$\overline{ X }= a +\frac{0+ d +2 d +\ldots+10 d }{11}$

$=a+5 d$

$\Rightarrow$ varience $=\frac{\Sigma\left(\bar{X}-x_{i}\right)^{2}}{11}$

$\Rightarrow 90 \times 11=\left(25 d^{2}+16 d^{2}+9 d^{2}+4 d^{2}\right) \times 2$

$\Rightarrow d =\pm 3 \Rightarrow d =3$

Standard 11

Mathematics

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