How to find the sum α=3+4+8+9+13+14+… upto 40 terms?

Split into odd-position terms (3,8,13,…) and even-position terms (4,9,14,…). Each is an AP with 20 terms and d=5. Sum of odd-position: 20/2×(6+95)=1010. Sum of even-position: 20/2×(8+95)=1030. α=2040.

What are the roots of x²+x−2=0?

x²+x−2=(x+2)(x−1)=0, so roots are x=1 and x=−2.

What is tanβ if (tanβ)²=1?

Since β∈(0,π/2), tanβ>0. So tanβ=1, giving β=π/4.

What is sin²β+3cos²β at β=π/4?

sin²(π/4)+3cos²(π/4)=1/2+3×(1/2)=1/2+3/2=2.

α=3+4+8+9+13+14+… upto 40 terms, (tanβ)^(α/1020) root of x²+x−2=0, find sin²β+3cos²β

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MathSequences & Series

Q MCQ Sequences & Trigonometry

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

A) $2$ B) $\dfrac{7}{4}$ C) $\dfrac{5}{2}$ D) $\dfrac{3}{2}$

✅ Correct Answer

A) 2

✓

Solution

Split the series into two interleaved APs

Series: $3,4,8,9,13,14,\ldots$ — the pattern is two consecutive terms increasing by 5 each pair.

Odd-position terms (1st, 3rd, 5th, …): $3,8,13,\ldots$ — AP with $a=3$, $d=5$, 20 terms.

Even-position terms (2nd, 4th, 6th, …): $4,9,14,\ldots$ — AP with $a=4$, $d=5$, 20 terms.

$S_{\text{odd}}=\dfrac{20}{2}[2(3)+19(5)]=10[6+95]=10\times101=1010$

$S_{\text{even}}=\dfrac{20}{2}[2(4)+19(5)]=10[8+95]=10\times103=1030$

$\alpha=1010+1030=\mathbf{2040}$

Find the exponent α/1020

$$\frac{\alpha}{1020}=\frac{2040}{1020}=2$$

So $(\tan\beta)^2$ is a root of $x^2+x-2=0$.

Find the roots of x² + x − 2 = 0

$x^2+x-2=(x+2)(x-1)=0$

$\Rightarrow\quad x=1\quad\text{or}\quad x=-2$

Since $(\tan\beta)^2\geq0$, only $(\tan\beta)^2=1$ is valid ($x=-2$ rejected).

Find β and compute the expression

$(\tan\beta)^2=1$ and $\beta\in(0,\pi/2)$, so $\tan\beta=1$, giving $\beta=\dfrac{\pi}{4}$.

$$\sin^2\beta+3\cos^2\beta=\frac{1}{2}+3\times\frac{1}{2}=\frac{1}{2}+\frac{3}{2}=\boxed{2}$$

📘 Key Concept

When a series alternates between two APs, split into odd-indexed and even-indexed terms. Each forms its own AP. Sum separately, then add. Here both sub-APs have $d=5$ and 20 terms each, making computation straightforward.

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Important Concepts

Interleaved AP Series Series like $a_1, a_1+1, a_1+d, a_1+d+1, \ldots$ consist of two APs with same common difference. Always split into odd and even positioned terms for efficient summation.

Sum of AP Formula $S_n=\dfrac{n}{2}[2a+(n-1)d]$ or $S_n=\dfrac{n}{2}[a+l]$ where $l$ is last term. For 20 terms with $a=3$, $d=5$: last term $=3+19\times5=98$, sum $=10(3+98)=1010$.

Rejecting Invalid Roots $(\tan\beta)^2$ must be non-negative. Root $x=-2$ is rejected because $(\tan\beta)^2\geq0$ always. Domain constraints eliminate extraneous roots.

Standard Trig Values at π/4 $\sin\frac{\pi}{4}=\cos\frac{\pi}{4}=\frac{1}{\sqrt{2}}$. So $\sin^2\frac{\pi}{4}=\cos^2\frac{\pi}{4}=\frac{1}{2}$. $\sin^2\beta+3\cos^2\beta=1+2\cos^2\beta=1+1=2$.

FAQs

Why split series into odd and even positions?

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The series $3,4,8,9,13,14,\ldots$ doesn't follow a single AP pattern. But odd-position terms $3,8,13,\ldots$ and even-position terms $4,9,14,\ldots$ each form clean APs with $d=5$, making the sum easy.

What is the last term of the odd-position AP?

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20th term $= a+(n-1)d = 3+19\times5 = 3+95 = 98$. Sum $= \frac{20}{2}(3+98)=10\times101=1010$.

Why is x=−2 rejected?

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$(\tan\beta)^2$ is a square of a real number, so it is always $\geq0$. The root $x=-2$ would require $(\tan\beta)^2=-2$, which is impossible for real $\beta$.

Can sin²β+3cos²β be simplified further?

⌄

Yes: $\sin^2\beta+3\cos^2\beta = \sin^2\beta+\cos^2\beta+2\cos^2\beta = 1+2\cos^2\beta$. At $\beta=\pi/4$: $1+2(1/2)=2$.

Is this from JEE Main 2026?

⌄

Yes, this question appeared in JEE Main 2026.

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