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

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

Statistics

Solution:

Mean = $\frac{101 + d(1 + 2 + 3 + .... + 100)}{101}$
$ = 1 + \frac{d \times 100 \times 101}{101 \times 2} = 1 + 50 d$
$\because $ Mean deviation from the mean = 255
$\Rightarrow \frac{1}{101} [\left|1- \left(1+ 50d\right)\right| + \left|\left(1+d\right) - \left(1+50d\right)\right|$
$ + \left|\left(1+2d\right) - \left(1+50d\right)\right| + ... +\left|\left(1+100d\right) - \left(1+50d\right)\right|] = 255 $
$ \Rightarrow 2d \left[ 1 + 2 + 3 + ....+50\right] = 101 \times255$
$ \Rightarrow 2d \times \frac{50 \times51}{2} = 101 \times 255$
$\Rightarrow d = \frac{101 \times255}{50 \times51} = 10.1$