If (a – b), (b – c), (c – a) are in G.P., then prove that (a + b + c)² = 3(ab + bc + ca)

Given as (a – b), (b – c), (c – a) are in G.P

∴

⇒ (b – c)² = (a – b)(c – a)

As we have to prove :(a + b + c)² = 3(ab + bc + ca) so we proceed as follows:

⇒ b² + c² – 2bc = ac – a² – bc + ab

⇒ a² + b² + c² = ac + ab + bc

Add 2(ac + ab + bc) to both sides:

⇒ a² + b² + c² + 2(ac + ab + bc) = ac + ab + bc + 2(ac + ab + bc)

⇒ (a + b + c)² = 3(ab + bc + ca)

Hence Proved.