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

$\begin{array}{rcl} & & 1, 1+\mathrm{d}, 1+2\mathrm{d}, ...... 1+100\mathrm{d}\\ \mathrm{Mean}& =& \frac{1+1+\mathrm{d}+1+2\mathrm{d}+.....+1+100\mathrm{d}}{101}\\ & =& 1+50\mathrm{d}\end{array}$

M.D. about mean $=\frac{\sum \left|{\mathrm{x}}_{\mathrm{i}}-\overline{\mathrm{x}}\right|}{\mathrm{n}}$

$$\begin{gathered} 255=\frac{2(\mathrm{d}+2\mathrm{d}+3\mathrm{d}+.....+50\mathrm{d})}{101} \\ 255=\frac{2\mathrm{d} \left(50\right) 51}{101 \left(2\right)} \Rightarrow \mathrm{d}=\frac{101}{10}=10.1 \end{gathered}$$