Let and be the roots of , and and be the roots of . If and , then is equal to [2025]

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Solution

Q.
Let $α$ and $β$ be the roots of $x^{2} + \sqrt{3} x - 16 = 0$ , and $γ$ and $δ$ be the roots of $x^{2} + 3 x - 1 = 0$ . If $P_{n} = α^{n} + β^{n}$ and $Q_{n} = γ^{n} + δ^{n}$ , then $\frac{P_{25} + \sqrt{3} P_{24}}{2 P_{23}} + \frac{Q_{25} - Q_{23}}{Q_{24}}$ is equal to [2025]

1 4

2 7

3 5

4 3

Ans.
(3)

We have, $x^{2} + 3 x - 1 = 0$ ... (i)

$\Rightarrow x^{2} - 1 = - 3 x$

$γ$ and $δ$ are the roots of equation (i)

Now, $Q_{n} = γ^{n} + δ^{n}$

$Q_{25} - Q_{23} = (γ^{25} - γ^{23}) + (δ^{25} - δ^{23})$

$= γ^{23} (γ^{2} - 1) + δ^{23} (δ^{2} - 1)$

   $= γ^{23} (- 3 γ) + δ^{23} (- 3 δ) = - 3 [γ^{24} + δ^{24}] = - 3 Q_{24}$

$\Rightarrow \frac{Q_{25} - Q_{23}}{Q_{24}} = (- 3)$

Similarly, $x^{2} + \sqrt{3} x - 16 = 0$ ... (ii)

$\Rightarrow x^{2} + \sqrt{3} x = 16$

$α$ and $β$ are the roots of equation (ii).

Now, $P_{n} = α^{n} + β^{n}$

$\Rightarrow P_{25} + \sqrt{3} P_{24} = (α^{25} + \sqrt{3} α^{24}) + (β^{25} + \sqrt{3} β^{24})$

   $= α^{23} (α^{2} + \sqrt{3} α) + β^{23} (β^{2} + \sqrt{3} β)$

   $= α^{23} (16) + 16 β^{23}$

$\Rightarrow \frac{P_{25} + \sqrt{3} P_{24}}{2 \cdot P_{23}} = \frac{16 (α^{23} + β^{23})}{2 (α^{23} + β^{23})} = 8$

Now, $\frac{P_{25} + \sqrt{3} P_{24}}{2 P_{23}} + \frac{(Q_{25} - Q_{23})}{Q_{24}} = 8 + (- 3) = 5$ .

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