JEE Mathematics 2025 Question with Solution

Given: $\alpha, \beta$ are roots of $x^2 + \sqrt{3}x - 16 = 0$ and $\gamma, \delta$ are roots of $x^2 + 3x - 1 = 0$.

Find: The value of $\frac{P_{25} + \sqrt{3}P_{24}}{2P_{23}} + \frac{Q_{25} - Q_{23}}{Q_{24}}$ where $P_n = \alpha^n + \beta^n$ and $Q_n = \gamma^n + \delta^n$.

For the first quadratic, since $\alpha, \beta$ are roots of

$$x^2 + \sqrt{3}x - 16 = 0$$

we get the recurrence relation

$$P_n + \sqrt{3}P_{n-1} - 16P_{n-2} = 0$$

Putting $n = 25$,

$$P_{25} + \sqrt{3}P_{24} - 16P_{23} = 0$$

So,

$$P_{25} + \sqrt{3}P_{24} = 16P_{23}$$

Hence,

$$\frac{P_{25} + \sqrt{3}P_{24}}{2P_{23}} = \frac{16P_{23}}{2P_{23}} = 8$$

Now for the second quadratic, since $\gamma, \delta$ are roots of

$$x^2 + 3x - 1 = 0$$

we have

$$\gamma^2 = -3\gamma + 1, \qquad \delta^2 = -3\delta + 1$$

Then

$$Q_{25} - Q_{23} = \gamma^{25} + \delta^{25} - \gamma^{23} - \delta^{23}$$ $$= \gamma^{23}(\gamma^2 - 1) + \delta^{23}(\delta^2 - 1)$$

Using $\gamma^2 - 1 = -3\gamma$ and $\delta^2 - 1 = -3\delta$,

$$Q_{25} - Q_{23} = -3\gamma^{24} - 3\delta^{24} = -3(\gamma^{24} + \delta^{24}) = -3Q_{24}$$

Therefore,

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

Now adding both parts,

$$\frac{P_{25} + \sqrt{3}P_{24}}{2P_{23}} + \frac{Q_{25} - Q_{23}}{Q_{24}} = 8 - 3 = 5$$

Therefore, the correct option is C.

Given: $P_n = \alpha^n + \beta^n$ for roots $\alpha, \beta$ of $x^2 + \sqrt{3}x - 16 = 0$, and $Q_n = \gamma^n + \delta^n$ for roots $\gamma, \delta$ of $x^2 + 3x - 1 = 0$.

Find: $\frac{P_{25} + \sqrt{3}P_{24}}{2P_{23}} + \frac{Q_{25} - Q_{23}}{Q_{24}}$.

If $r$ is a root of $x^2 + ax + b = 0$, then

$$r^2 = -ar - b$$

and the sequence of power sums satisfies

$$S_n + aS_{n-1} + bS_{n-2} = 0$$

for $S_n = r_1^n + r_2^n$.

Applying this to $x^2 + \sqrt{3}x - 16 = 0$,

$$P_n + \sqrt{3}P_{n-1} - 16P_{n-2} = 0$$

At $n = 25$,

$$P_{25} + \sqrt{3}P_{24} = 16P_{23}$$

Thus,

$$\frac{P_{25} + \sqrt{3}P_{24}}{2P_{23}} = \frac{16P_{23}}{2P_{23}} = 8$$

Next, for $x^2 + 3x - 1 = 0$,

$$\gamma^2 = -3\gamma + 1, \qquad \delta^2 = -3\delta + 1$$

So,

$$\gamma^2 - 1 = -3\gamma, \qquad \delta^2 - 1 = -3\delta$$

Now,

$$\begin{aligned} Q_{25} - Q_{23} &= (\gamma^{25} + \delta^{25}) - (\gamma^{23} + \delta^{23}) \\ &= \gamma^{23}(\gamma^2 - 1) + \delta^{23}(\delta^2 - 1) \\ &= \gamma^{23}(-3\gamma) + \delta^{23}(-3\delta) \\ &= -3\gamma^{24} - 3\delta^{24} \\ &= -3Q_{24} \end{aligned}$$

Therefore,

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

Finally,

$$8 + (-3) = 5$$

So the expression is $5$, hence the correct option is C.