Question:
If the variance of the terms in an increasing A.P., $\mathrm{b}_{1}, \mathrm{~b}_{2}, \mathrm{~b}_{3}, \ldots \mathrm{b}_{11}$ is 90 , then the common difference of this A.P, is.__________.
Solution:
Let a be the first term and $d$ be the common difference of the given A.P. Where $d>0$
$\bar{X}=a+\frac{0+d+2 d+\ldots+10 d}{11}$
$=a+5 d$
$\Rightarrow$ varience $=\frac{\Sigma\left(\bar{X}-x_{i}\right)^{2}}{11}$
$\Rightarrow 90 \times 11=\left(25 \mathrm{~d}^{2}+16 \mathrm{~d}^{2}+9 \mathrm{~d}^{2}+4 \mathrm{~d}^{2}\right) \times 2$
$\Rightarrow \mathrm{d}=\pm 3 \Rightarrow \mathrm{d}=3$