Discussion
The product of first four terms of a G.P. is \(\frac{1}{1024}\). Then the product of second and third terms is,
Answer: D
The product of the second and third terms is \(\frac{1}{32}\).
In a G.P. \(a, ar, ar^{2}, ar^{3}\), terms equidistant from the ends have equal products: \(T_{1}T_{4} = a^{2}r^{3} = T_{2}T_{3}\). So the product of all four is \((T_{2}T_{3})^{2}\):
\[ \begin{aligned} (T_{2}T_{3})^{2} &= \frac{1}{1024} \\ T_{2}T_{3} &= \frac{1}{32} \end{aligned} \]
(The positive root is taken, as the options are all positive.) The key step is the square root: \(\sqrt{1024} = 32\). \(\frac{1}{64}\), \(\frac{1}{16}\) and \(\frac{1}{128}\) are other powers of 2 whose squares are not \(\frac{1}{1024}\), and \(\frac{1}{28}\) is not a power of 2 at all.
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