A) \[n(n+1)d\]
B) \[\frac{n(n+1)d}{2n+1}\]
C) \[\frac{n(n+1)d}{2n}\]
D) \[\frac{n(n-1)d}{2n+1}\]
Correct Answer: B
Solution :
[b] The mean of the series \[\bar{X}=\frac{1}{2n+1}\{a+(a+d)+(a+2d)+...+(a+2nd)\}\] \[=\frac{1}{2n+1}\left\{ \frac{2n+1}{2}(2a+2nd) \right\}=a+nd\] Therefore, mean deviation from mean \[\frac{1}{2n+1}\sum\limits_{r=0}^{2n}{\left| (a+rd)-(a+nd) \right|}\] \[=\frac{1}{2n+1}\sum\limits_{r=0}^{2n}{\left| r-n \right|d}\] \[=\frac{[2(1+2+...+n)+0]d}{2n+1}=\frac{n(n+1)d}{2n+1}\]
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