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

Step 1. Define the terms of the A.P :

Let the first term and the common difference of the A.P. be $a$ and $d$. And as it is an increasing A.P. so $d>0$. The terms of the A.P. are defined as:

$\begin{array}{rcl}{b}_1& =& a,\\ b_2& =& a+d,\\ b_3& =& a+2d,\\ & & ...\\ b_{11}& =& a+10d\end{array}$

Step 2. Find the mean of the terms

The mean of the $11$ terms is given as:

$\begin{array}{rcl}\overline{X}& =& \frac{b_1+b_2+b_3+...+b_{11}}{11}\\ & =& \frac{a+(a+d)+(a+2d)+...+(a+10d)}{11}\\ & =& \frac{11a+(1+2+...+10)d}{11}\\ & =& \frac{11a}{11}+\frac{55d}{11}\\ & =& a+5d\end{array}$

Step 3. Find the common difference of the A.P.

Now the variance of the terms is $90$, so

$\begin{array}{rcl}\mathrm{Variance}& =& \frac{{\displaystyle \sum _{i=1}^{11}}{\left(\overline{X}-b_i\right)}^2}{11}\\ & \Rightarrow & 90=\frac{{\left(\overline{X}-b_1\right)}^2+{\left(\overline{X}-b_2\right)}^2+...{\left(\overline{X}-b_{11}\right)}^2}{11}\\ & \Rightarrow & 90\times 11={\left(a+5d-a\right)}^2+{\left(a+5d-(a+d)\right)}^2+....+{\left(a+5d-\left(a+10d\right)\right)}^2\\ & \Rightarrow & 990={\left(5d\right)}^2+{\left(4d\right)}^2+{\left(3d\right)}^2+{\left(2d\right)}^2+{\left(d\right)}^2+{\left(-d\right)}^2+{\left(-2d\right)}^2+{\left(-3d\right)}^2+{\left(-4d\right)}^2+{\left(-5d\right)}^2\\ & \Rightarrow & 990=\left(25+16+9+4+1+1+4+9+16+25\right)d^2\\ & \Rightarrow & 990=110d^2\\ & \Rightarrow & \frac{990}{110}=d^2\\ & \Rightarrow & 9=d^2\\ & \Rightarrow & \pm \sqrt{9}=d\\ & \Rightarrow & \pm 3=d\end{array}$

But the value of $d$ must be greater than $0$, so $d=-3$ is rejected and the solution $d=3$ is accepted.

Hence, the common difference of the A.P. is $d=3$.