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

Let $X$ denote the mean of the given numbers.

Then,

$\bar{X} = \frac{1 + (1 + d) + (1 + 2 d) + \dots + (1 + 100 d)}{101}$

$\Rightarrow \bar{X} = \frac{\frac{101}{2} {1 + (1 + 100 d))\}}{101} = 1 + 50 d$

$∴$ Mean deviation $= \frac{1}{101} \{\sum_{r = 0}^{100} | (1 + r d) - (1 + 50 d) |\}$

$\Rightarrow$ Mean deviation $= \frac{1}{101} \sum_{r = 0}^{100} | r - 50 |$

$\Rightarrow$ Mean deviation $= \frac{d}{101} \times 2 \sum_{r = 1}^{50} r$

$\Rightarrow$ Mean deviation $= \frac{2 d}{101} \times \frac{50 \times 51}{2} = \frac{50 \times 51}{101} d .$

It is given that the mean deviation is $255 .$

$∴ 255 = \frac{50 \times 51}{101} d \Rightarrow d = 10.1$

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