Question

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

Answer 1

Let $b_1=a,b_2=a+d,b_3=a+2d...,b_{11}=a+10d$
Var $\left(x\right)=\frac{\sum b_i^2}{11}-{\left(\frac{\sum b_i}{11}\right)}^2$

$\therefore 90=\left\{\frac{11a^2+385d^2+110ad}{11}\right\}-{\left\{\frac{11a+55d}{11}\right\}}^2$
$90=\{a^2+35d^2+10ad\}-{\{a+5d\}}^2$
$90=a^2+35d^2+10ad-a^2-25d^2-10ad$
$90=10d^2$
$9=d^2$
$\Rightarrow d=3$