Let be the distinct roots of the equation x 2 - ( t 2 - 5 t + 6 ) x + 1 = 0 , t ? R and . Then the minimum value of a 2023 + a 2025 a 2024 is [2024]

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Solution

Q.
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 [2024]

1 $- 1 / 2$

2 $1 / 2$

3 $1 / 4$

4 $- 1 / 4$

Ans.
(4)

Newton's Theorem says that for a quadratic equation $a x^{2} + b x + c = 0$ if $a$ and $b$ are its roots and $a_{n} = α^{n} \pm β^{n},$ then $a a_{n + 1} + b a_{n} + c a_{n - 1} = 0 .$

So, by Newton's theorem, we have

$a_{2025} - (t^{2} - 5 t + 6) a_{2024} + a_{2023} = 0$

$\Rightarrow t^{2} - 5 t + 6 = \frac{a_{2025} + a_{2023}}{a_{2024}}$

Now, $t^{2} - 5 t + 6 = {(t - \frac{5}{2})}^{2} - \frac{1}{4}$

Minimum value $= - \frac{1}{4}$

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