Let, α, β be the distinct roots of the equation x^2 - (t^2 - 5t + 6)x + 1 = 0, t ∈ R and an = α^n + β^n

Question

Let, α, β be the distinct roots of the equation

\(\mathrm{x}^{2}-\left(\mathrm{t}^{2}-5 \mathrm{t}+6\right) \mathrm{x}+1=0, \mathrm{t} \in \mathrm{R}\, and \,\mathrm{a}_{\mathrm{n}}=\alpha^{\mathrm{n}}+\beta^{\mathrm{n}}\)

Then the minimum value of \(\frac{a_{2023}+a_{2025}}{a_{2024}}\) is

(1) \( \frac{1}{4}\)

(2) \( \frac{-1}{2}\)

(3) \( \frac{-1}{4}\)

(4) \( \frac{1}{2}\)

Answer 1

Correct option is (3) \(\frac{-1}{4}\)

by newton's theorem

\( a_{n+2}-\left(t^{2}-5 t+6\right) a_{n+1}+a_{n}=0\)

\(\therefore \mathrm{a}_{2025}+\mathrm{a}_{2023}=\left(\mathrm{t}^{2}-5 \mathrm{t}+6\right) \mathrm{a}_{2024}\)

\( \therefore \frac{\mathrm{a}_{2025}+\mathrm{a}_{2023}}{\mathrm{a}_{2024}}=\mathrm{t}^{2}-5 \mathrm{t}+6\)

\(\because \mathrm{t}^{2}-5 \mathrm{t}+6=\left(\mathrm{t}-\frac{5}{2}\right)^{2}-\frac{1}{4}\)

∴ minimum value \(=-\frac{1}{4}\)