The mean deviation from the mean of the A.P. $a,a+d,a+2d,\dots \dots \dots ,a+2\mathrm{nd}$ is.

The mean of the series

$\begin{array}{l}\overline{\mathrm{X}}=\frac{1}{2n+1}\{a+(a+d)+(a+2d)+\dots ..+(a+2nd\left)\right\}\\ =\frac{1}{2n+1}\left\{\frac{2n+1}{2}(2a+2nd)\right\}=a+nd\end{array}$

Therefore, mean deviation from mean

$\begin{array}{l}\frac{1}{2n+1}\sum _{r=0}^{2n} \left|\right(a+rd)-(a+nd\left)\right|=\frac{1}{2n+1}\sum _{r=0}^{2n} |r-n|d\\ =\frac{\left[2\right(1+2+\dots +n)+0]d}{2n+1}=\frac{n(n+1)d}{2n+1}\end{array}$