If $r_1,r_2,r_3$, in triangle be in H.P. then the sides are:

A.P. G.P. H.P. None of these

A.P.

$\frac{1}{r_3}-\frac{1}{r_2}=\frac{1}{r_2}-\frac{1}{r_1}$ $⇒\frac{1}{r_1}+\frac{1}{r_3}=\frac{2}{r_2}$ so $\frac{s-a}{Δ}+\frac{s-c}{Δ}=\frac{2(s-b)}{Δ}$ $⇒2s-(a+c)=2s-2b$ $⇒a+c=2b$