If the mean deviation of the numbers 1, 1 + d, , 1 + 100d from...

Mean $\overset{-}{x} = \dfrac{1}{101}\left[1 + \left(1 + d\right) + \left(1 + 2d\right)\right] \ldots . \left(1 + 100d\right)$

$= \dfrac{1}{101} \times \dfrac{101}{2}\left[1 + \left(1 + 100d\right)\right] = 1 + 50d$

Mean deviation from mean

$= \dfrac{1}{101}\left\{\left|1 - \left(1 + 50d\right)\right| + \left|\left(1 + d\right) - \left(1 + 50d\right)\right| \ldots . \left|\left[1 + 100d\right] - \left(1 + 50d\right)\right|\right\}$

$= \dfrac{2\left|d\right|}{101}\left(1 + 2 + 3 \ldots . + 50\right)$

$= \dfrac{2\left|d\right|}{101}\dfrac{\left(50\right)\left(50 + 1\right)}{2}$

$= \dfrac{2550}{101}\left|d\right| = 255$

$⇒ \left|d\right| = 10 .1$