Discussion
Let \(t_{1}, t_{2}, t_{3}, ..., t_{n}\) be in G.P. Then \(\left(\dfrac{t_{4}}{t_{2}}\right)^{3}\) is equal to
Answer: B
\(\left(\frac{t_{4}}{t_{2}}\right)^{3}\) equals \(\left(\frac{t_{7}}{t_{4}}\right)^{2}\).
In a G.P. with common ratio \(r\), \(\frac{t_{m}}{t_{n}} = r^{m - n}\). So
\[ \begin{aligned} \left(\frac{t_{4}}{t_{2}}\right)^{3} &= (r^{2})^{3} = r^{6} \\ \left(\frac{t_{7}}{t_{4}}\right)^{2} &= (r^{3})^{2} = r^{6} \end{aligned} \]
The others give different powers: \(\left(\frac{t_{7}}{t_{3}}\right)^{2} = r^{8}\), \(\left(\frac{t_{6}}{t_{4}}\right)^{2} = r^{4}\), \(\left(\frac{t_{7}}{t_{2}}\right)^{2} = r^{10}\) and \(\left(\frac{t_{9}}{t_{8}}\right)^{2} = r^{2}\).
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