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If the mean deviation of number $1,1+d, 1+2 d, \ldots ., 1+100 d$ from their mean is 255 , then the $d$ is equal to
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Verified Answer
The correct answer is:
$10.1$
$10.1$
$$
\begin{aligned}
& \text { Mean }(\bar{x})=\frac{\text { sum of quantities }}{n}=\frac{\frac{n}{2}(a+1)}{\not n}=\frac{1}{2}[1+1+100 d]=1+50 d \\
& \text { M.D. }=\frac{1}{n} \sum\left|x_i-\bar{x}\right| \Rightarrow 255=\frac{1}{101}[50 d+49 d+48 d+\ldots .+d+0+d+\ldots \ldots+50 d]=\frac{2 d}{101}\left[\frac{50 \times 51}{2}\right] \\
& \Rightarrow d=\frac{255 \times 101}{50 \times 51}=10.1
\end{aligned}
$$
\begin{aligned}
& \text { Mean }(\bar{x})=\frac{\text { sum of quantities }}{n}=\frac{\frac{n}{2}(a+1)}{\not n}=\frac{1}{2}[1+1+100 d]=1+50 d \\
& \text { M.D. }=\frac{1}{n} \sum\left|x_i-\bar{x}\right| \Rightarrow 255=\frac{1}{101}[50 d+49 d+48 d+\ldots .+d+0+d+\ldots \ldots+50 d]=\frac{2 d}{101}\left[\frac{50 \times 51}{2}\right] \\
& \Rightarrow d=\frac{255 \times 101}{50 \times 51}=10.1
\end{aligned}
$$
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