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

$\frac{1}{r_3}-\frac{1}{r_2}=\frac{1}{r_2}-\frac{1}{r_1}$

$⇒\frac{1}{r_1}+\frac{1}{r_3}=\frac{2}{r_2}$

so $\frac{s-a}{Δ}+\frac{s-c}{Δ}=\frac{2(s-b)}{Δ}$

$⇒2s-(a+c)=2s-2b$

$⇒a+c=2b$