Question

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

Answer 1

Ans--n(n1)d/2n1

2 : the standard deviation of data

X 1 a a^2........ a^n

F nC0 nC1 nC2.......nCn

Ans----((1a^2)/2)^n - ((1a)/2)^n



____________ different answer method

Given series is
a, a+d, a+2d, ............, a+2nd
meanx¯=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