Let α = 3+4+8+9+13+14+... upto 40 terms. If (tan β) α is a root of the equation x²+x-2=0, β∈(0,π/2), then sin²β+3cos²β is equal to:

Mathematics · Sequence and Series

JEE Main 2026 — 8 April, Evening Shift — Question 24

Let $α = 3+4+8+9+13+14+...$ upto 40 terms. If $(\tan \beta)^{\alpha}$ } is a root of the equation $x²+x-2=0, β∈(0,π/2),$ then $sin²β+3cos²β$ is equal to:

Option A:
$2$
Correct
Option B:
$46119$
Option C:
$46058$
Option D:
$46056$

Answer: A

Step-by-step solution

$\mathrm{a}=(3+8+13 \ldots$ upto 20 terms $)+(4+9+14+$ ... upto 20 terms) $=\frac{20}{2}[6+19 \times 5]+\frac{20}{2}(8+19 \times 5)$ $=2040$ $(\tan \beta)^{\frac{2040}{1020}}=\tan ^{2} \beta$ $\mathrm{x}^{2}+\mathrm{x}-2=0<{ }_{-2}^{1}$ $\Rightarrow \tan ^{2} \beta=1 ; \sin ^{2} \beta=\frac{1}{2} \& \cos ^{2} \beta=\frac{1}{2}$ $\sin ^{2} \beta+3 \cos ^{2} \beta=\frac{1}{2}+\frac{3}{2}=\frac{4}{2}=2$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 8 April, Evening Shift
Subject: Mathematics
Chapter: Sequence and Series
Topic: Arithmetic Progression