Question

The mean deviation of the series $a,a+d,a+2d+......,a+(2n-1)d,a+2nd$ about the mean is

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

Answer 1

Answer: (B)

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

Mean $\overline{x}=\frac{\left(a\right)+(a+d)+(a+2d)+...+(a+2nd)}{2n+1}$
where $\overline{x}=\frac{\sum x_i}{n}$
Then, $\frac{\frac{2n+1}{2}(a+a+2nd)}{2n+1}$
From an A.P. $S_n=\frac{n}{2}(a+l)$ which gives $\overline{x}=a+nd$
The series being a,a+d,a+2d,...,a+(n-1)d,a+nd,a+(n+1)d,...,a+2nd
Mean deviation from the mean
= $\frac{1}{N}\sum f_i[x_i-\overline{x}]$
= $\frac{1}{2n+1}\sum [x_i-a-nd]$
= $\frac{1}{2n+1}[nd+(n-1)d+(n-2)d+...+d+0+d+...+nd]$
= $\frac{2d}{2n+1}[n+(n-1)+(n-2)+...1]$
= $\frac{2d}{2n+1}.\frac{n(n+1)}{2}=\frac{n(n+1)d}{2n+1}$