The first term of a Geometric Progression (GP) is \(\frac{3}{4}\), If the product of the second and third terms of the sequence is 972, find its common ratio.
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Correct Answer: Option D
Explanation:
Recall, the terms of a GP is given by \(T_{n} = ar^{n - 1}\)
Given, \(T_{1} = a = \frac{3}{4}\) and \(T_{2} \times T_{3} = ar \times ar^{2} = 972\)
\(\implies \frac{3}{4}r \times \frac{3}{4}r^{2} = \frac{9}{16}r^{3} = 972\)
\(r^{3} = \frac{972 \times 16}{9} = 1728 \implies r = \sqrt[3]{1728} = 12\)
Recall, the terms of a GP is given by \(T_{n} = ar^{n - 1}\)
Given, \(T_{1} = a = \frac{3}{4}\) and \(T_{2} \times T_{3} = ar \times ar^{2} = 972\)
\(\implies \frac{3}{4}r \times \frac{3}{4}r^{2} = \frac{9}{16}r^{3} = 972\)
\(r^{3} = \frac{972 \times 16}{9} = 1728 \implies r = \sqrt[3]{1728} = 12\)