Let, α, β be the distinct roots of the equation x 2 - (t 2 -5 t+6 ) x+1=0, t in R and a n =α n +β n . Then the minimum value of frac a 2023 +a 2025 a 2024 is

Mathematics · Quadratic Equations

JEE Main 2024 — 6 April, Shift 1 — Question 11

Let, $\alpha, \beta$ be the distinct roots of the equation $x^{2}-\left(t^{2}-5 t+6\right) x+1=0, t \in R$ and $a_{n}=\alpha^{n}+\beta^{n}$ .

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

by newton's theorem

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

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

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

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

$\therefore$ minimum value $=-\frac{1}{4}$

Answer key and solution verified before publishing.

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