Let α, β be the distinct roots of the equation x²-(t²-5 t+6) x+1=0, t ∈ ℝ and aₙ=αⁿ+βⁿ. Then the minimum value of (a₂₀₂₃+a₂₀₂₅)/a₂₀₂₄ is

Solution

Maths sir · 4 steps

Each root satisfies the quadratic, so multiply it by a power of that root and add the two copies.

1. Name the coefficient. Let s = t² - 5t + 6. The equation is x² - sx + 1 = 0, so each root satisfies x² = sx - 1.
2. Build the recurrence. Multiply α² = sα - 1 by α²⁰²³ to get α²⁰²⁵ = sα²⁰²⁴ - α²⁰²³. Do the same for β and add the two results.

   a₂₀₂₅ + a₂₀₂₃ = s a₂₀₂₄ ⇒ (a₂₀₂₃+a₂₀₂₅)/a₂₀₂₄ = s = t² - 5t + 6

3. Minimise the quadratic. Complete the square: t² - 5t + 6 = (t - 5/2)² - 1/4. The least value is -1/4, at t = 5/2. At this value the discriminant s² - 4 is negative, so the roots are complex conjugates, which are still distinct.
4. So. The minimum value of the expression is -1/4, which is option (b).

The idea to keep. For aₙ = αⁿ + βⁿ with α, β roots of x² - px + q = 0, use aₙ₊₂ = p aₙ₊₁ - q aₙ. Huge powers then reduce to the coefficients.

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