Question

If x + 2ay + az = 0, x + 3by + bz = 0,

x + 4 cy + cz = 0 have a non-zero solution, then a, b, c –

• A. Are in GP
• B. Are in HP
• C. Satisfy a + 2b + 3c = 0
• D. Are in AP

Answer 1

Answer: (B)

Are in HP

Answer 2

Q ∆ = $\left| \begin{matrix} 1 & 2a & a \\ 1 & 3b & b \\ 1 & 4c & c \end{matrix} \right|$ = 0

⇒ 1 (– bc) – 2a (c – b) + a (4c – 3b) = 0

⇒ –bc –2ac + 2ab + 4ac – 3ab = 0

⇒ 2ac = ab + bc ⇒ b = $\frac{2ac}{a + c}$ ∴ a,b,c are in HP.