If the variance of the terms of an increasing A.P.b1,b2,…,b11 is 90, then the common difference of this A.P. is

A
3
B
-3
C
9
D
-9

Solution:

Let $d$ be the common difference of the AP. then,

$\begin{array}{l} b_{i} = b_{1} + (i - 1) d i = 1, 2, \dots, 11 . \\ \Rightarrow Var (b_{1}, b_{2}, \dots, b_{11}) = Var (0 d, 1 d, 2 d, \dots, 10 d) \\ \Rightarrow 90 = Var (d, 2 d, 3 d, \dots, 10 d) \\ \Rightarrow 90 = d^{2} Var (1, 2, 3, \dots, 10) \end{array}$

$\begin{array}{l} \Rightarrow 90 = d^{2} (\frac{10^{2} - 1}{12}) \\ \Rightarrow 90 = 10 d^{2} \Rightarrow d = 3 [\begin{matrix} ∵ b_{1}, b_{2}, \dots, b_{11} is an increasing A.P. \\ ∴ d > 0 \end{matrix}] \end{array}$

If the variance of the terms of an increasing A.P.
$b_{1}, b_{2}, \dots, b_{11}$ is 90, then the common difference of this A.P. is

Solution:

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