Discussion
Let \(a, \frac{3}{4}, ar^{2}, ar^{3}, \ldots\) be in G.P. where \(r \gt 0\) If the product of first four terms of the G.P. is \(\frac{3^{6}}{4^{5}}\) then \(a\) is equal to
Answer: D
\(a = \frac{1}{2}\).
The second term is \(ar = \frac{3}{4}\). The product of the first four terms is \(a \cdot ar \cdot ar^{2} \cdot ar^{3} = a^{4}r^{6}\), which can be written using \(ar\):
\[ \begin{aligned} a^{4}r^{6} &= \frac{(ar)^{6}}{a^{2}} = \frac{3^{6}}{4^{6}a^{2}} \\ \frac{3^{6}}{4^{6}a^{2}} &= \frac{3^{6}}{4^{5}} \Rightarrow a^{2} = \frac{1}{4} \end{aligned} \]
With \(r \gt 0\) and \(ar \gt 0\), \(a = \frac{1}{2}\) (then \(r = \frac{3}{2}\)). The other values do not satisfy the product condition.
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