If the mean deviation of the numbers $1,1+\text{d},1+2\text{d}\mathrm{........1}+100\text{d}$ from their mean is 255 then d = ____

$A.M=\frac{1+1+d+1+2d+1+3d+\mathrm{....1}+100d}{101}$
$=1+d\frac{100\times 101}{2\times 101}=1+50d$
$M.D=255$
$\begin{array}{l}\frac{\left(50d+49d+\mathrm{...}+d+0+d+2d+\mathrm{...}+50d\right)}{101}\\ =255\end{array}$
$\frac{\frac{2\times 50\times 51}{2}}{101}=255$
$d=\frac{101}{10}=10.1$