Question

Mean deviation of the series $a,a+d,a+2d,...,a+2nd$ from its mean is

• A. $\frac{(n+1)d}{(2n+1)}$
• B. $\frac{nd}{2n+1}$
• C. $\frac{(2n+1)d}{n(n+1)}$
• 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 $a,a+d,a+2d,.......,a+2nd$

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

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

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

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

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

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

$=a+nd$

Now mean devaition from mean

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

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

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

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