Question

If 4a² + 9b² + 16c² = 2(3ab + 6bc + 4ca), where a, b, c are non-zero numbers, then a, b, c are in

• A. A.P.
• B. G.P.
• C. H.P
• D. None of these

Answer 1

Answer: (C)

H.P

Answer 2

The given equation 4a² + 9b² + 16c² = 2(3ab + 6bc + 4ca)

⇒ (2a – 3b)² + (3b – 4c)² + (4c – 2a)² = 0

⇒ 2a = 3b = 4c

⇒ $\frac { a } { 1 / 2 } = \frac { b } { 1 / 3 } = \frac { c } { 1 / 4 }$ ⇒ a, b, c are in H.P