The mean deviation from the mean of the AP  a, a + d , a+2d , ……..a+ nd is

The mean of the seriesa,a+d,a+2d,........…,a+2ndX¯=12n+1[a+(a+d)+(a+2d)+…+(a+2nd)]=12n+12n+12(a+a+2nd)=a+ndTherefore, mean deviation from mean =12n+1∑r=02n |(a+rd)−(a+nd)|=12n+1∑r=02n |r−n|d=12n+1⋅2d(1+2+…+n)=n(n+1)2n+1d