Question

The mean deviation of the series a, a + d, a + 2d, ..., a + 2n from its mean is
(a) $\frac{(n+1)d}{2n+1}$

(b) $\frac{nd}{2n+1}$

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

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

Answer 1

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

$x_i$	$\left|x_i-\overline{X}\right|=\left|x_i-\left(a+nd\right)\right|$
a	nd
a + d	(n-1)d
a + 2d	(n-2)d
a + 3d	(n-3)d
:	:
:	:
a + (n - 1)d	d
a + nd	0
a + (n+1)d	d
:	:
:	:
a + 2nd	nd
$\sum _{}{x}_i=\left(2n+1\right)\left(a+nd\right)$	$\sum _{}\left|x_i-\overline{X}\right|=n\left(n+1\right)d$

$$\begin{gathered} \mathrm{There}\mathrm{are}2n+1\mathrm{terms}. \\ \Rightarrow N=2n+1 \\ \sum _{}{x}_i=a+a+d+a+2d+a+3d+...+a+2nd \\ =(2n+1)a+d(1+2+3+...+2n)\left[a+a+a+...(2n+1)\mathrm{times}=\left(2n+1\right)a\right] \\ =(2n+1)a+\frac{2n\left(2n+1\right)d}{2}\left[\mathrm{Sum}\mathrm{of}\mathrm{the}\mathrm{first}n\mathrm{natural}\mathrm{numbers}\mathrm{is}\frac{n(n+1)}{2},\mathrm{but}\mathrm{here}\mathrm{we}\mathrm{are}\mathrm{considering}\mathrm{the}\mathrm{first}2n\mathrm{numbers}.\right] \\ =\left(2n+1\right)a+\left(2n+1\right)nd \\ =\left(2n+1\right)\left(a+nd\right) \\ \overline{X}=\frac{\left(2n+1\right)\left(a+nd\right)}{\left(2n+1\right)} \\ =a+nd \\ \sum _{}\left|x_i-\overline{X}\right|\mathbf{=}nd+(n-1)d+(n-2)d+...+d+0+d+2d+3d+...+nd \\ =d\left(n+\left(n-1\right)+\left(n-2\right)+...+1\right)+0+d\left(1+2+3+...+n\right) \\ =\frac{dn\left(n+1\right)}{2}+\frac{dn\left(n+1\right)}{2}\left\{\because 1+2+3+...+n=\frac{n\left(n+1\right)}{2}\right\} \\ =n(n+1)d \\ \mathrm{Mean}\mathrm{deviation}\mathrm{about}\mathrm{the}\mathrm{mean}=\frac{\sum _{}\left|x_i-\overline{X}\right|\mathbf{}}{N} \\ =\frac{n(n+1)d}{\left(2n+1\right)} \end{gathered}$$