Question

The mean deviation of the series a, a + d, a + 2d, ......, a + 2nd from its mean is

Answer 1

The given Series is a, a+d, a+2d, a+3d, ..., a+2ndSo, its first term=a
and common difference=d and mth term=a+2nd

let no. of terms be m.
So, am=a+(m-1)d⇒a+2nd=
a+(m-1)d⇒2nd=(m-1)d⇒2n=(m-1)⇒m=2n+1So, nowSm=m2(a+am)⇒Sm=2n+12(a+a+2nd)⇒Sm=2n+12(2a+2nd)⇒Sm=2n+122(a+nd)⇒Sm=(2n+1)(a+nd)
So, Arithmetic Mean=Smm=(2n+1)(a+nd)(2n+1)=a+nd

Hence Arithmetic Mean of the Series is a+nd.


Regards
 

Answer 2

Given series is
a, a+d, a+2d, ............, a+2nd
mean(x⎯⎯)=a+a+d+a+2d+........+a+2nd2n+1              =a(2n+1)+(d+2d+.......+2nd)2n+1              =a(2n+1)+d(1+2+3+....+2n)2n+1              =a(2n+1)+d2n(2n+1)22n+1            =a(2n+1)+dn(2n+1)2n+1            =(2n+1)(a+nd)2n+1            =a+ndNow deviation from mean i.e xi−xnd, (n−1)d, (n−2)d,     ......., 0, (n−1)d, (n−2)d, ndNow mean deviation from mean=nd+ (n−1)d+(n−2)d+ .......+0+d+2d+....+(n−1)d+(n−2)d+ nd2n+1=2d(1+2+3+....(n−1)+(n−2)+n)2n+1=2(n(n+1)2)d2n+1=n(n+1)d2n+1


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