JEE Main 2025 Apr 3 Shift 1, Mathematics Q3: Arithmetic, Geometric And Harmonic Progressions

JEE Main2025Apr 3, Shift 1Mathematics

Let $α$ and $β$ be the roots of $x^{2} + 3 x - 16 = 0$ , 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 $\frac{P _{25} + 3 P _{24}}{2 P _{23}} + \frac{Q _{25} - Q _{23}}{Q _{24}}$ is equal to

Solution

If $α, β$ satisfy $x^{2} + 3 x - 16 = 0$ , then for each root $α^{n} + 3 α^{n - 1} - 16 α^{n - 2} = 0$ , and similarly for $β$ . Adding,

$P_{n} + 3 P_{n - 1} - 16 P_{n - 2} = 0.$

At $n = 25$ : $P_{25} + 3 P_{24} = 16 P_{23}$ , so

$\frac{P _{25} + 3 P _{24}}{2 P _{23}} = \frac{16}{2} = 8.$

For $γ, δ$ satisfying $x^{2} + 3 x - 1 = 0$ , we get $γ^{2} = 1 - 3 γ$ , so $γ^{25} - γ^{23} = γ^{23} (γ^{2} - 1) = γ^{23} (- 3 γ) = - 3 γ^{24}$ . Same for $δ$ . Adding,

$Q_{25} - Q_{23} = - 3 Q_{24} \Rightarrow \frac{Q _{25} - Q _{23}}{Q _{24}} = - 3.$

Sum: $8 + (- 3) = 5.$

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JEE Main 2025 Apr 3 Shift 1, Mathematics Q3: Arithmetic, Geometric And Harmonic Progressions

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