Question

The mean deviation from the mean of the series $a,a+d,a+2d,....a+2nd$, 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)

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

Clearly given observation is in A.P, and have $2n+1$ terms.
$\therefore$ Mean $=\frac{1}{2}\{a+a+2nd\}=a+nd$
And thus mean deviation about mean $=\frac{1}{2n+1}\sum |x-\overline{x}|$ $=\frac{1}{2n+1}.2d(1+2+3\cdots +n)$ $=\frac{n(n+1)d}{2n+1}$