If (a – b), (b – c), (c – a) are in G.P., then prove that (a + b

1.	If (a – b), (b – c), (c – a) are in G.P., then prove that (a + b + c)2 = 3(ab + bc + ca)
Answer» Given as (a – b), (b – c), (c – a) are in G.P \(\therefore\) \(\frac{b-c}{a-b}\) = \(\frac{c-a}{b-c}\) = common ratio ⇒ (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.

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

Answer»

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

\(\therefore\) \(\frac{b-c}{a-b}\) = \(\frac{c-a}{b-c}\) = common ratio

⇒ (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.