Question

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

• A. $20$
• B. $10.1$
• C. $20.2$
• D. $10$

Answer 1

The correct option is B $10.1$

Mean
$=\frac{101+d(1+2+3+\cdots +100)}{101}=1+\frac{d\times 100\times 101}{101\times 2}=1+50d$

Now,
$\text{Mean deviation from the mean =255}\Rightarrow \frac{[|1-(1+50d)|+|1+d-(1+50d)|+|1+2d-(1+50d)|+\cdots +|1+100d-(1+50d\left)|\right]}{101}=255\Rightarrow 2|d|[1+2+3+\dots 50]=101\times 255\Rightarrow 2|d|\times \frac{50\times 51}{2}=101\times 255\Rightarrow |d|=\frac{101\times 255}{50\times 51}\therefore |d|=10.1$