2. Suppose that $IG\parallel BC$ (I=incenter, G=centroid).
Let $r$ be the inradius, $h_a$ the height from the vertex A, $S$ the area of triangle and $p$ the semiperimeter.
Then we have $r=\frac{h_a}{3}=\frac{2S}{3a}$
But $r=\frac{S}{p}\Rightarrow\frac{2S}{3a}=\frac{S}{p}\ Rightarrow\frac{1}{3a}=\frac{1}{a+b+c}\Rightarrow a=\frac{b+c}{2}$ and that means that b, a, c (or c, a, b) are in arithmetic progression.