3
4
6
12
Correct answer is D
Recall, the terms of a GP is given by \(T_{n} = ar^{n - 1}\)
Given, \(T_{1} = a = \frac{3}{4}\) and \(T_{2} \times T_{3} = ar \times ar^{2} = 972\)
\(\implies \frac{3}{4}r \times \frac{3}{4}r^{2} = \frac{9}{16}r^{3} = 972\)
\(r^{3} = \frac{972 \times 16}{9} = 1728 \implies r = \sqrt[3]{1728} = 12\)