Question

If the mean deviation of the number $1,1+d,1+2d,....1+100d$ from their mean is $255$ then $d$ is equal to $(d>0)$

• A. $10.0$
• B. $20.0$
• C. $10.1$
• D. $20.2$

Answer 1

The correct option is C $10.1$

$1,1+d,1+2d,....1+100d$ are the observations
So, $n$(no. of observation)$=101$
Mean, $\overline{x}=\frac{1+1+1d+1+2d+1+3d+.....+1+100d}{101}=\frac{101+(d+2d+3d+..100d)}{101}=\frac{101+d(1+2+3+..100)}{101}=\frac{101+d\left(\frac{100\times 101}{2}\right)}{101}=1+50d$
Men deviation = $\frac{\sum _{i=1}^n|x_i-\overline{x}|}{n}$
put the value of $x_i$ and $\overline{x}$ we get mean deviation as
$\frac{|1-(1+50d)|+|1+d-(1+50d)|+|1+2d-(1+50d)|+...+|1+100d-(1+50d)|}{101}=255\Rightarrow \frac{|-50d|+|-49d|+|-48d|+...+|-d|+0+|d|+|2d|+...+|50d|}{101}=255\Rightarrow \frac{2(d+2d+3d+......+50d)}{101}=255\Rightarrow \frac{2d(50\times 51)}{101}=255\Rightarrow d=\frac{101}{10}$