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

A
10.0
B
20.0
C
10.1
D
20.2

Solution:

$\begin{array}{l} (\bar{x}) = \frac{Sum of quantities}{n} = \frac{\frac{n}{2} (a + l)}{n} \\ = \frac{1}{2} (1 + 1 + 100 d) = 1 + 50 d \\ ∴ MD = \frac{1}{n} \sum |x_{i} - \bar{x}| \\ \Rightarrow 255 = \frac{1}{101} (50 d + 49 d + 48 d + \dots + d + 0 + d + \dots + 50 d) \\ = \frac{2 d}{101} (\frac{50 \times 51}{2}) \\ \Rightarrow d = \frac{255 \times 101}{50 \times 51} = 10 .1 \end{array}$

Solution:

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