A) (a) \[\frac{n(n+1).d}{2n+1}\]
B) (b) \[n(n+1).d\]
C) (c) \[\frac{n(n-1).d}{2n+1}\]
D) (d) \[\frac{n(n+1).d}{2n}\]
Correct Answer: A
Solution :
[a] \[\because \overline{x}=\frac{\frac{2n+1}{2}(a+a+2nd)}{2n+1}\] \[=a+nd\sum{\left| x-\overline{x} \right|}=1+2+3+......n\] \[=2nd\frac{n(n+1)d}{2n+1}=n(n+1).d\] \[\therefore \]Mean deviation \[=\frac{n(n+1)d}{2n+1}\] Hence, option [a]is correct.
You need to login to perform this action.
You will be redirected in
3 sec
