Question

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

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

Answer 1

Answer: (B)

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

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

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

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

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

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

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

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

$=a+nd$

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 mean

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

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

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

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