Question

Show that the mean deviation from the mean of the A.P $a,a+d,a+2d,....a+2nd$ is independent of the common difference of A.P

Answer 1

mean$=\frac{\frac{(2n+1)}{2}(2a+(2nd)}{2n+1}=a+nd$

mean deviation from mean is $\frac{{\Sigma }_1^{2n+1}(a+(r-1)d-a-nd)}{2n+1}=\frac{{\Sigma }_1^{2n+1}(r-n-1)d}{2n+1}=0$

Mean deviation from mean is independent of the common difference of A.P