The given series is an AP where 1st term = a , common difference = d and number of terms = 2n+1
Hence sum of the series is S = 2n/2[2a+(2n+1-1)d]
Hence mean is m = S/2n=a+nd
So absolute value of deviations of the terms are
nd, (n-1)d,(n-2d), ....2d,..d,0, d, 2d, .....nd
Hence sum of the deviations is D= 2d(1+2+3+4+...+n)= 2dxn(n+1)/2=n(n+1)d
Hence the average deviation Dav= D/2n = (n+1)d/2