Question

The mean deviation from mean of the observation $a,a+d,a+2d,.....a+2nd$ is

• A. $\frac{n(n+1)d^2}{3}$
• B. $\frac{n(n+1)}{2}{d}^2$
• C. $a+\frac{n(n+1)d^2}{2}$
• D. $\frac{n(n+1)d}{(2n+1)}$

Answer 1

The correct option is D $\frac{n(n+1)d}{(2n+1)}$

Given series is $a,a+d,a+2d,...,a+2nd$

$mean\left(\overline{x}\right)=\frac{a+a+d+a+2d+....+a+2nd}{2n+1}$

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

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

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

$=a+nd$

Now, deviation from mean i.e., $x_i-\overline{x}$

$nd,(n-1)d,(n-2)d,....0,(n-1)d,(n-2)d,nd$

Now, mean deviation from meam

$=\frac{nd+(n-1)d+(n-2)d+....+0+d+2d+....d(n-1)d+(n-2)d+nd}{2n+1}$

$=\frac{2d(1+2+3+...(n-1)+(n-2)+n)}{2n+1}$

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