If (a – b), (b – c), (c – a) are in GP then prove that (a + b + c)2 = 3(ab + bc + ca).
To prove: (a + b + c)2 = 3(ab + bc + ca).
Given: (a – b), (b – c), (c – a) are in GP
Formula used: When a,b,c are in GP, b2 = ac
As, (a – b), (b – c), (c – a) are in GP
⇒ (b – c)2 = (a – b) (c – a)
⇒ b2 -2cb + c2 = ac – a2 – bc + ab
⇒ a2 + b2 + c2 – bc – ac – ab = 0
Adding 3(ab + bc + ac) both side
⇒ a2 + b2 + c2 – bc – ac – ab + 3(ab + bc + ac) = 3(ab + bc + ac)
⇒ a2 + b2 + c2 + 2bc + 2ac + 2ab = 3(ab + bc + ac)
⇒ (a + b + c)2 = 3(ab + bc + ac)
Hence Proved