AP: \(a,\,a+d,\,a+2d,\,\dots,\,a+2nd\) has \(2n+1\) terms. The mean is the middle term \(a+nd\).

Mean deviation from the mean (MD): average of absolute deviations from \(a+nd\).

Symmetry gives pairs at distances \(kd\) for \(k=1,\dots,n\). Hence

$$\text{MD}=\frac{1}{2n+1}\Big[2\sum_{k=1}^{n} kd\Big] =\frac{2d}{2n+1}\cdot\frac{n(n+1)}{2} =\frac{d\,n(n+1)}{2n+1}.$$

Answer: \(\displaystyle \boxed{\frac{d\,n(n+1)}{2n+1}}\).