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.