Let $\alpha=3+4+8+9+13+14+\ldots$ upto 40 terms. If $(\tan \beta)^{\frac{\alpha}{1020}}$ is a root of the equation $x^{2}+x-2=0, \beta \in\left(0, \frac{\pi}{2}\right)$, then $\sin ^{2} \beta+3 \cos ^{2} \beta$ is equal to :

JEE Main · 2026 · 08 Apr 2026 · Shift 2 · Sequences and Series

Let

α = 3 + 4 + 8 + 9 + 13 + 14 + \dots

upto 40 terms. If

(tan β)^{\frac{α}{1020}}

is a root of the equation

x^{2} + x - 2 = 0, β \in (0, \frac{π}{2})

, then

sin^{2} β + 3 cos^{2} β

is equal to :

Answer: (A)

The forty terms split into two arithmetic progressions. The terms in odd positions,

3, 8, 13, \dots

, form an AP with first term

3

and common difference

5

; the terms in even positions,

4, 9, 14, \dots

, form an AP with first term

4

and common difference

5

. Each has

20

terms.

odd \sum = \frac{20}{2} (2 \cdot 3 + 19 \cdot 5) = 10 (6 + 95) = 1010, even \sum = \frac{20}{2} (2 \cdot 4 + 19 \cdot 5) = 10 (8 + 95) = 1030,

α = 1010 + 1030 = 2040

and

\frac{α}{1020} = 2

. Hence

(tan β)^{2}

is a root of

x^{2} + x - 2 = (x + 2) (x - 1) = 0

, giving

x = 1

x = - 2

. Since

β \in (0, \frac{π}{2})

tan β > 0

, so

tan^{2} β = 1

and

β = \frac{π}{4}

. Then

sin^{2} β = cos^{2} β = \frac{1}{2}

, and

sin^{2} β + 3 cos^{2} β = \frac{1}{2} + 3 \cdot \frac{1}{2} = 2.

JEE Main · 2026 · 08 Apr 2026 · Shift 2 · Single correct option