If $\alpha, \beta$ and $\gamma$ are three consecutive terms of a non-constant G.P. such that the equations $\alpha x^{2}+2 \beta x+\gamma=0$ and $x^{2}+x-1=0$ have a common root, then $\alpha(\beta+\gamma)$ is equal to :

JEE Main · 2019 · 12 Apr 2019 · Shift 2 · Sequences and Series

α, β

and

γ

are three consecutive terms of a non-constant G.P. such that the equations

α x^{2} + 2 β x + γ = 0

and

x^{2} + x - 1 = 0

have a common root, then

α (β + γ)

is equal to :

Answer: (D)

Write the G.P. as

α, β = α r, γ = α r^{2}

. Then

α x^{2} + 2 β x + γ = 0

becomes

x^{2} + 2 r x + r^{2} = 0

, a perfect square with the double root

x = - r

. This must be the root shared with

x^{2} + x - 1 = 0

, so

(- r)^{2} + (- r) - 1 = 0

, giving

r^{2} = r + 1

. Now

α (β + γ) = α^{2} r (1 + r) = α^{2} (r + r^{2}) = α^{2} (2 r + 1)

, and since

r^{3} = r \cdot r^{2} = r (r + 1) = 2 r + 1

, this equals

α^{2} r^{3} = (α r) (α r^{2}) = β γ

. So

α (β + γ) = β γ

JEE Main · 2019 · 12 Apr 2019 · Shift 2 · Single correct option