Question

If $\cos \theta, \sqrt{2}\sin \theta$ and $\sqrt{3}\tan \theta$, where $0 < \theta < \frac{\pi}{2}$, are the second, third and fourth terms of a geometric series, respectively, then the first term of the geometric series is

• A. $\frac{1}{2\sqrt{6}}$
• B. $\frac{3}{2\sqrt{6}}$
• C. $\frac{1}{\sqrt{6}}$
• D. $\frac{3}{\sqrt{6}}$
• E. $\frac{2}{\sqrt{6}}$

Hint:

In GP problems involving trigonometry, equate the common ratio $r = T_n / T_{n-1}$ to find the angle first. This usually simplifies the radical expressions significantly.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
In a Geometric Progression (GP), the ratio between any two consecutive terms is constant, known as the common ratio $r$. If $a$ is the first term, then $T_n = a r^{n-1}$.
Key Formula or Approach:
Given $T_2 = ar = \cos \theta$, $T_3 = ar^2 = \sqrt{2}\sin \theta$, and $T_4 = ar^3 = \sqrt{3}\tan \theta$.
We use the property $r = \frac{T_3}{T_2} = \frac{T_4}{T_3}$.

Step 2: Detailed Explanation:
1. Equate the ratios:
\[ \frac{\sqrt{2}\sin \theta}{\cos \theta} = \frac{\sqrt{3}\tan \theta}{\sqrt{2}\sin \theta} \]
\[ \sqrt{2}\tan \theta = \frac{\sqrt{3}\frac{\sin \theta}{\cos \theta}}{\sqrt{2}\sin \theta} \]
\[ \sqrt{2}\tan \theta = \frac{\sqrt{3}}{\sqrt{2}\cos \theta} \]
\[ 2 \frac{\sin \theta}{\cos \theta} = \frac{\sqrt{3}}{\cos \theta} \]
Since $0 < \theta < \frac{\pi}{2}$, $\cos \theta \neq 0$:
\[ 2\sin \theta = \sqrt{3} \implies \sin \theta = \frac{\sqrt{3}}{2} \implies \theta = 60^\circ \]
2. Find the common ratio $r$:
\[ r = \frac{\sqrt{2}\sin 60^\circ}{\cos 60^\circ} = \sqrt{2} \tan 60^\circ = \sqrt{2} \cdot \sqrt{3} = \sqrt{6} \]
3. Find the first term $a$:
\[ ar = \cos 60^\circ = \frac{1}{2} \]
\[ a(\sqrt{6}) = \frac{1}{2} \implies a = \frac{1}{2\sqrt{6}} \]

Step 3: Final Answer:
The first term is $\frac{1}{2\sqrt{6}}$.