Let Pₙ=αⁿ+βⁿ, n ∈ N. If P₁₀=123, P₉=76, P₈=47 and P₁=1, then the quadratic equation having roots 1/α and 1/β is :

Solution

Maths sir · 4 steps

Power sums of two roots follow a recurrence, and one step of it gives you αβ directly.

1. Recurrence for P. Each of α and β satisfies t² = (α+β)t - αβ. Multiply by tⁿ⁻² and add the two versions.

   Pₙ = (α+β) Pₙ₋₁ - αβ Pₙ₋₂

2. Find the product. P₁ = α+β = 1. Put n = 10: 123 = 1 × 76 - αβ × 47, so 47 αβ = -47 and αβ = -1.
3. Reciprocal roots. Sum: 1/α+1/β = (α+β)/αβ = 1/-1 = -1. Product: 1/αβ = -1.
4. So. The equation is x² - (-1)x + (-1) = 0, that is x² + x - 1 = 0, which is option (a).

The idea to keep. For Pₙ = αⁿ + βⁿ, always use Pₙ = (α+β)Pₙ₋₁ - αβ Pₙ₋₂. For reciprocal roots, the sum is (α+β)/αβ and the product is 1/αβ.

Answer from the NTA final answer key. Solution by PrepWiser.