Question

If the mean deviation about mean $1,1+d,1+2d,....1+100d$ from their mean is $255$, then the $d$ is equal to

• A. $10$
• B. $20$
• C. $10.1$
• D. $20.2$

Answer 1

The correct option is C $10.1$

Clearly given observation is in A.P, and have $101$ terms.
$\therefore$ Mean $=\frac{1}{2}\{1+1+100d\}=1+50d$
And thus mean deviation about mean $=\frac{1}{101}\sum |x-\overline{x}|$ $=\frac{1}{101}.2d(1+2+3\cdots +50)$ $=\frac{50\left(51\right)d}{101}=255$ (given)
$\Rightarrow d=10.1$