Given that \(\sqrt{6}, 3\sqrt{2}, 3\sqrt{6}, 9\sqrt{2},...\) are the first four terms of an exponential sequence (G.P), find in its simplest form the 8th term.
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Correct Answer: Option C
Explanation:
\(T_{n} = ar^{n - 1}\) (Geometric progression)
\(a = \sqrt{6}, r = \frac{T_{2}}{T_{1}} = \frac{3\sqrt{2}}{\sqrt{6}} \)
\(r = \frac{\sqrt{18}}{\sqrt{6}} = \sqrt{3}\)
\(\therefore T_{8} = (\sqrt{6})(\sqrt{3})^{8 - 1} \)
= \((\sqrt{6})(27\sqrt{3}) = 27\sqrt{18} = 81\sqrt{2}\)
\(T_{n} = ar^{n - 1}\) (Geometric progression)
\(a = \sqrt{6}, r = \frac{T_{2}}{T_{1}} = \frac{3\sqrt{2}}{\sqrt{6}} \)
\(r = \frac{\sqrt{18}}{\sqrt{6}} = \sqrt{3}\)
\(\therefore T_{8} = (\sqrt{6})(\sqrt{3})^{8 - 1} \)
= \((\sqrt{6})(27\sqrt{3}) = 27\sqrt{18} = 81\sqrt{2}\)