If the variance of the terms in an increasing A.P., $b_{1}, b_{2}, b_{3}, \ldots ., b_{11}$ is 90 ,

Question

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

Professor PC Thomas Classes · Answer

(3): Given, $b_{1}, b_{2}, b_{3}, \ldots, b_{11}$ are in A.P. and variance of $\left(b_{1}, b_{2}, \ldots, b_{11}ight)=90$\ $\Rightarrow \quad \frac{\sum_{i=1}^{11} b_{i}^{2}}{11}-\left(\frac{\sum_{i=1}^{11} b_{i}}{11}ight)^{2}=90$\ $\Rightarrow \frac{b_{1}^{2}+b_{2}^{2}+\ldots+b_{11}^{2}}{11}-\left(\frac{b_{1}+b_{2}+\ldots+b_{11}}{11}ight)^{2}=90$\ Let $d$ be the common difference of A.P. then\ $\frac{b_{1}^{2}+\left(b_{1}+dight)^{2}+\ldots+\left(b_{1}+10 dight)^{2}}{11}$\ $\begin{aligned} & -\left[\frac{b_{1}+\left(b_{1}+dight)+\ldots+\left(b_{1}+10 dight)}{11}ight]^{2}=90 \ \Rightarrow \quad & \frac{1 \mathrm{i} b_{1}^{2}+d^{2}\left(1^{2}+2^{2}+\ldots+10^{2}ight)}{+2 b_{1} d(1+2+\ldots+10)} \ 11 & \frac{1}{2}\end{aligned}$\ $-\left(\frac{11 b_{1}+d(1+2+\ldots+10)}{11}ight)^{2}=90$\ $\Rightarrow \quad b_{1}^{2}+\frac{d^{2}}{11} \cdot \frac{10(11)(21)}{6}+\frac{2 b_{1} d}{11} \cdot \frac{10(11)}{2}$ $$ -\left[b_{1}+\frac{d}{11} \cdot \frac{10(11)}{2}ight]^{2}=90 $$ $\Rightarrow \quad b_{1}^{2}+d^{2}\left(\frac{210}{6}ight)+10 b_{1} d$ $$ -\left[b_{1}^{2}+(5 d)^{2}+10 b_{1} dight]=90 $$ $\Rightarrow \quad 35 d^{2}-25 d^{2}=90 \Rightarrow 10 d^{2}=90$\ $\Rightarrow d^{2}=9 \Rightarrow d=3$\ $[\because$ Given that A.P. is increasing. So, $d$ can't be negative]\