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

# If the mean deviation of the numbers  from their mean is $255,$ then $d$ is equal to

1. A

$10.0$

2. B

$20.0$

3. C

$10.1$

4. D

$20.2$

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### Solution:

Let $\overline{)X}$ denote the mean of the given numbers.

Then,

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

Mean deviation $=\frac{1}{101}\left\{\sum _{r=0}^{100} |\left(1+rd\right)-\left(1+50d\right)|\right\}$

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

Mean deviation $=\frac{d}{101}×2\sum _{r=1}^{50}r$

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

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

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