In a ΔABC, $\tan\frac{A}{2}=\frac{5}{6},\,\tan\frac{C}{2}=\frac{2}{5}$ then:

a,c,b are in A.P a,b,c are in A.P b,a,c are in A.P a,b,c are in G.P

a,b,c are in A.P

$\tan\frac{A}{2}.\tan\frac{C}{2}=\frac{s-b}{s}=\frac{5}{6}.\frac{2}{5}=\frac{1}{3}$ ⇒ 3s - 3b = s ⇒ 2s = 3b ⇒ a + c = 2b ⇒ a, b, c as in AP