If the mean deviation of the data $1,1+d$, $1+2 d, \ldots, 1+100 d,(d0)$ from their mean is 255 , then ' $d$ ' is equal to

Question

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

Aicharya · Accepted Answer

Mean $\bar{x}=\frac{	ext { Sum of quantities }}{n}$ $=\frac{\frac{n}{2}(a+l)}{n} \quad[l=$ last team $]$ $\begin{aligned} & =\frac{1}{2}(a+l)=\frac{1}{2}(1+1+100 d) \ & =1+50 d\end{aligned}$ $\mathrm{MD}=\frac{1}{n} \Sigma\left|x_i-\bar{x}ight|$ $\begin{aligned} \Rightarrow 255=\frac{1}{101}[50 d & +49 d+48 d+\ldots \ & +d+0+d+\ldots .+50 d]\end{aligned}$ $\Rightarrow 255=\frac{2 d}{101}\left[\frac{50 	imes 51}{2}ight]$ $d=\frac{255 	imes 101}{50 	imes 51}=10.1$

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 '…

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