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Mathematics · Quadratic Equations

JEE Main 2024 — 27 January, Shift 2 — Question 10

If α,β\alpha, \beta are the roots of the equation, x2−x−1=0x^{2}-x-1=0 and Sn=2023αn+2024βnS_{n}=2023 \alpha^{n}+2024 \beta^{n}, then

  1. Option A:

    2 S12=S11+S102 \mathrm{~S}_{12}=\mathrm{S}_{11}+\mathrm{S}_{10}

  2. Option B:

    S12=S11+S10\mathrm{S}_{12}=\mathrm{S}_{11}+\mathrm{S}_{10}

    Correct
  3. Option C:

    2 S11=S12+S102 \mathrm{~S}_{11}=\mathrm{S}_{12}+\mathrm{S}_{10}

  4. Option D:

    S11=S10+S12\mathrm{S}_{11}=\mathrm{S}_{10}+\mathrm{S}_{12}

Answer: B

Step-by-step solution

x2−x−1=0x^{2}-x-1=0

Sn=2023αn+2024βnS_{n}=2023 \alpha^{n}+2024 \beta^{n}

Sn−1+Sn−2=2023αn−1+2024βn−1+2023αn−2+2024βn−2\mathrm{S}_{\mathrm{n}-1}+\mathrm{S}_{\mathrm{n}-2}=2023 \alpha^{\mathrm{n}-1}+2024 \beta^{\mathrm{n}-1}+2023 \alpha^{\mathrm{n}-2}+2024 \beta^{\mathrm{n}-2}

=2023αn−2[1+α]+2024βn−2[1+β]=2023 \alpha^{\mathrm{n}-2}[1+\alpha]+2024 \beta^{\mathrm{n}-2}[1+\beta]

=2023αn−2[α2]+2024βn−2[β2]=2023 \alpha^{\mathrm{n}-2}\left[\alpha^{2}\right]+2024 \beta^{\mathrm{n}-2}\left[\beta^{2}\right]

Sn−1+Sn−2=2023αn+2024βn=Sn\mathrm{S}_{\mathrm{n}-1}+\mathrm{S}_{\mathrm{n}-2}=2023 \alpha^{\mathrm{n}}+2024 \beta^{\mathrm{n}}=\mathrm{S}_{n}

Sn−1+Sn−2=Sn\mathrm{S}_{\mathrm{n}-1}+\mathrm{S}_{\mathrm{n}-2}=\mathrm{S}_{\mathrm{n}}

Put n=12\mathrm{n}=12

S11+S10=S12\mathrm{S}_{11}+\mathrm{S}_{10}=\mathrm{S}_{12}

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
27 January, Shift 2
Subject
Mathematics
Chapter
Quadratic Equations
Topic
Theory of Quadratic Equations
If α, β are the roots of the equation, x 2 -x-1=0 and S n =2023 α n… | JEE Main 2024 PYQ with Solution · DhiX AI