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

The correct option is C

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

$\begin{array}{cc}{x}_i& |x_i-\overline{X}|=|x_i-(a+nd)|\\ a& nd\\ a+d& (n-1)d\\ a+2d& (n-2)d\\ a+3d& (n-3)d\\ :& :\\ :& :\\ a+(n-1)d& d\\ a+nd& 0\\ a+(n+1)d& d\\ :& :\\ :& :\\ a+2nd& nd\\ \sum x_i=(2n+1)(a+nd)& \sum |x_i-\overline{x}|=n(n+1)d\end{array}$

There are 2n+1 terms

$\Rightarrow N=2n+1$

$\sum x_i=a+a+d+a+2d+a+3d+....+a+2nd=(2n+1)a+d(1+2+3+...+2n)$

[a+a+a+.... (2n+1) times = (2n+1)a]

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

[Sum of the first n natural numbers is \( \frac{n(n+1)}{2}, but here we are considering the fir

= (2n+1)a+ (2n+1)nd = (2n+1)(a+nd)

$\overline{X}=\frac{(2n+1)(a+nd)}{(2n+1)}=a+nd$

$\sum |x_i-\overline{X}|=nd+(n-1)d+(n-2)d+....d+0+d+2d+3d+....+nd$

= d (n+ (n-1)+(n-2)+....+1)+0+d (1+2+3+....+n)

= $\frac{dn(n+1)}{2}+\frac{dn(n+1)}{2}$ $\{\because 1+2+3+...+n=\frac{n(n+1)}{2}\}$

= n(n+1) d

Mean deviation about the mean = $\frac{\sum |x_i-\overline{X}|}{N}=\frac{n(n+1)d}{(2n+1)}$