Question

The mean deviation from the mean of the series a a d a 2d a 2nd is

Answer 1

The number of terms of given A.P is (2n+1),

Mean, x¯=a+a+2nd/2=a+nd

xᵣ=a+rd,where r=0,1,......,2n

Now xᵣ-x‾=(a+rd)-(a+nd)=(r-n)d being considered postiive

2n  

\sum \,|xᵣ-x‾|=

r=0

=  2n  

d\sum \, |r-n|

r=0

=2n  

d[\sum \,{(n-r)+(r-n)}]=d{(1+2+.....+n)+(1+....+n)} =2dn(n+1)/2=n(n+1)d

r=0

Therefore Mean absolute deviation =n(n+1)d/2n+1