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If the mean deviation of the data $1,1+d$, $1+2 d, \ldots, 1+100 d,(d>0)$ from their mean is 255 , then ' $d$ ' is equal to
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The correct answer is:
$10.1$
Mean $\bar{x}=\frac{\text { Sum of quantities }}{n}$
$=\frac{\frac{n}{2}(a+l)}{n} \quad[l=$ last team $]$
$\begin{aligned} & =\frac{1}{2}(a+l)=\frac{1}{2}(1+1+100 d) \\ & =1+50 d\end{aligned}$
$\mathrm{MD}=\frac{1}{n} \Sigma\left|x_i-\bar{x}\right|$
$\begin{aligned} \Rightarrow 255=\frac{1}{101}[50 d & +49 d+48 d+\ldots \\ & +d+0+d+\ldots .+50 d]\end{aligned}$
$\Rightarrow 255=\frac{2 d}{101}\left[\frac{50 \times 51}{2}\right]$
$d=\frac{255 \times 101}{50 \times 51}=10.1$
$=\frac{\frac{n}{2}(a+l)}{n} \quad[l=$ last team $]$
$\begin{aligned} & =\frac{1}{2}(a+l)=\frac{1}{2}(1+1+100 d) \\ & =1+50 d\end{aligned}$
$\mathrm{MD}=\frac{1}{n} \Sigma\left|x_i-\bar{x}\right|$
$\begin{aligned} \Rightarrow 255=\frac{1}{101}[50 d & +49 d+48 d+\ldots \\ & +d+0+d+\ldots .+50 d]\end{aligned}$
$\Rightarrow 255=\frac{2 d}{101}\left[\frac{50 \times 51}{2}\right]$
$d=\frac{255 \times 101}{50 \times 51}=10.1$
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