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The mean deviation from the mean of the series \((a),(a+d),(a+2 d), \ldots \ldots \ldots . .,(a+2 n d)\) is
Options:
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Verified Answer
The correct answer is:
\(\frac{n(n+1) d}{2 n+1}\)
The mean of given series \(a, a+d, a+2 d, \ldots \ldots, a+2 n d\) is
\(\frac{\frac{2 n+1}{2}[2 a+(2 n) d]}{2 n+1}=(a+n d)=m\) (let)
So, the mean deviation \(=\frac{1}{2 n+1} \sum_{i=1}^{2 n+1}\left|x_i-m\right|\)
\(\begin{aligned}
& =\frac{1}{2 n+1}[|n d|+|n d-d|+|n d-2 d|+\ldots .+|n d-2 n d|] \\
& =\frac{2 d}{2 n+1}[n+(n-1)+(n-2)+\ldots .+1]=\frac{n(n+1) d}{2 n+1}
\end{aligned}\)
Hence, option (b) is correct
\(\frac{\frac{2 n+1}{2}[2 a+(2 n) d]}{2 n+1}=(a+n d)=m\) (let)
So, the mean deviation \(=\frac{1}{2 n+1} \sum_{i=1}^{2 n+1}\left|x_i-m\right|\)
\(\begin{aligned}
& =\frac{1}{2 n+1}[|n d|+|n d-d|+|n d-2 d|+\ldots .+|n d-2 n d|] \\
& =\frac{2 d}{2 n+1}[n+(n-1)+(n-2)+\ldots .+1]=\frac{n(n+1) d}{2 n+1}
\end{aligned}\)
Hence, option (b) is correct
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