Discussion
If \(\cos\theta\), \(\sqrt{2}\sin\theta\) and \(\sqrt{3}\tan\theta\), where \(0 \lt \theta \lt \frac{\pi}{2}\), are the second, third and fourth terms of a geometric series, respectively, then the first term of the geometric series is
Answer: A
The first term is \(\frac{1}{2\sqrt{6}}\).
In a geometric series the square of the middle of three consecutive terms equals the product of its neighbours:
\[ \begin{aligned} (\sqrt{2}\sin\theta)^{2} &= \cos\theta \cdot \sqrt{3}\tan\theta \\ 2\sin^{2}\theta &= \sqrt{3}\sin\theta \\ \sin\theta &= \tfrac{\sqrt{3}}{2}, \quad \theta = \tfrac{\pi}{3} \end{aligned} \]
So the second, third and fourth terms are \(\frac{1}{2}\), \(\frac{\sqrt{6}}{2}\) and 3, with common ratio \(\sqrt{6}\).
\[ \begin{aligned} \text{first term} &= \frac{1/2}{\sqrt{6}} \\ &= \frac{1}{2\sqrt{6}} \end{aligned} \]
The option \(\frac{1}{\sqrt{6}}\) forgets the factor \(\frac{1}{2}\) in \(\cos\frac{\pi}{3}\).
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