A) \[\frac{n(n+1)d}{2n-1}\]
B) \[\frac{n(n+1)d}{2n+1}\]
C) \[\frac{n(n-1)d}{2n+1}\]
D) \[\frac{(n+1)d}{2}\]
E) \[\frac{n(n-1)d}{2n-1}\]
Correct Answer: B
Solution :
The mean of the series\[a,a+d,....,\text{ }a+2nd\]is \[\overline{x}=\frac{1}{2n+1}[a+a+d+a+2d+....+a+2nd]\] \[=\frac{1}{2n+1}\left[ \frac{2n+1}{2}(a+a+2nd) \right]=a+nd\] \[\therefore \]Mean deviation from mean \[=\frac{1}{2n+1}\sum\limits_{r=0}^{2n}{|(a+rd)-(a+nd)|}\] \[=\frac{1}{2n+1}\sum\limits_{r=0}^{2n}{(r-n)}d\] \[=\frac{1}{2n+1}2d(1+2+....+n)=\frac{n(n+1)}{2n+1}d\]
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