Question

If the variance of the terms in an increasing $A.P.,b_1,b_2,b_3,\dots b_{11}$ is $90,$ then the common difference of this $A.P$ . is

Answer 1

Answer: 3.00

Let a be the first term and $d$ be the common difference of the given $A.P$. Where $d>0$
$\bar{X}=a+\frac{0+d+2d+\dots +10d}{11}$
$=a+5d$
$\Rightarrow \text{ varience }=\frac{\sum {\left(\bar{X}-x_i\right)}^2}{11}$
$\Rightarrow 90\times 11=\left(25d^2+16d^2+9d^2+4d^2\right)\times 2$
$\Rightarrow d=\pm 3\Rightarrow d=3$