If a, b, c are in

Given:

a,b,c are in AP

∴ 2b = a + c …… (i)

b,c,d are in GP;

⇒ c² = bd …… (ii)

1/c, 1/d, 1/e are in AP;

⇒

⇒ …(iii)

From the above substituting for b & d in (ii) above,

⇒

⇒ c(c + e) = (a + c) e

⇒ c² + ce = ae + ce

⇒ c^{2 =} ae

Thus a, c, e are in GP