Question

For the arithmetic progression, $a,(a+d),(a+2d),(a+3d),...,(a+2nd),$the mean deviation from the mean is

• A. $\left(\frac{n(n+1)}{(2n–1)}\right)d$
• B. $\left(\frac{n(n+1)}{(2n+1)}\right)d$
• C. $\left(\frac{n(n–1)}{(2n+1)}\right)d$
• D. $\frac{(n+1)d}{2}$
• E. $\left(\frac{n(n–1)}{(2n–1)}\right)d$

Answer 1

The correct option is B

$\left(\frac{n(n+1)}{(2n+1)}\right)d$

The explanation for the correct option:

Step-1. Calculate the mean:

Given,

Arithmetic Progression $a,(a+d),(a+2d),(a+3d),...,(a+2nd),$

Total number of terms $=2n+1$

First-term $=a$

Last term $=a+2nd$

Common difference $=d$

Sum $\begin{array}{rcl}S& =& \frac{\left(2n+1\right)\left(a+a+2nd\right)}{2}\mathbf{[}\mathbf{\because }\mathbf{}\mathbf{\text{'}}\mathbf{S}\mathbf{\text{'}}\mathbf{}\mathbf{=}\frac{\mathbf{n}\mathbf{(}\mathbf{a}\mathbf{}\mathbf{+}\mathbf{l}\mathbf{)}}{\mathbf{2}}\mathbf{]}\end{array}$

$\begin{array}{rcl}& =& \frac{\left(2n+1\right)\left(2a+2nd\right)}{2}\\ & =& \left(2n+1\right)\left(a+nd\right)\end{array}$

Mean $\begin{array}{rcl}& =& \frac{\left(2\mathrm{n}+1\right)\left(\mathrm{a}+\mathrm{nd}\right)}{\left(2\mathrm{n}+1\right)}\left[\mathbf{\because }\mathbf{Mean}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{S}}{(2n+1)}\right]\end{array}$

$\begin{array}{rcl}& =& \left(\mathrm{a}+\mathrm{nd}\right)\end{array}$

Step-2. Calculate Mean deviation from mean:

Mean deviation$\begin{array}{rcl}& =& \frac{1}{\left(2n+1\right)}\sum _{\mathrm{m}=0}^{2\mathrm{n}}|{\mathrm{x}}_{\mathrm{m}}-\mathrm{Mean}|\mathbf{[}\mathbf{\because }\mathbf{Mean}\mathbf{}\mathbf{deviation}\mathbf{}\mathbf{=}\frac{\mathbf{1}}{n}\mathbf{}\mathbf{}\mathbf{\sum }_{\mathbf{i}\mathbf{=}\mathbf{1}}^{\mathbf{n}}\mathbf{|}{\mathbf{x}}_{\mathbf{i}}\mathbf{-}\mathbf{Mean}\mathbf{|}\mathbf{]}\end{array}$

$\begin{array}{rcl}& =& \frac{1}{\left(2n+1\right)}\sum _{\mathrm{m}=0}^{\mathrm{n}}\left(nd+(n-1)d+(n-2)d+...+d+0+d+...+nd\right)\end{array}$ $\left[\because a-a-\mathrm{nd}=\mathrm{nd},a+d-a-\mathrm{nd}=(1-n)d....\right]$

$\begin{array}{rcl}& =& \frac{2d}{\left(2n+1\right)}\left(n+(n-1)+(n-2)+...1\right)\\ & =& \frac{2d}{\left(2n+1\right)}\frac{\left(n\right)\left(n+1\right)}{2}\\ & =& \frac{\mathbf{\left(}\mathbf{n}\mathbf{\right)}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{d}}{(2n+1)}\end{array}$

Hence, Option(B) is the correct answer.