\(27\sqrt{2}\)
\(27\sqrt{6}\)
\(81\sqrt{2}\)
\(81\sqrt{6}\)
Correct answer is C
\(T_{n} = ar^{n - 1}\) (Geometric progression)
\(a = \sqrt{6}, r = \frac{T_{2}}{T_{1}} = \frac{3\sqrt{2}}{\sqrt{6}} \)
\(r = \frac{\sqrt{18}}{\sqrt{6}} = \sqrt{3}\)
\(\therefore T_{8} = (\sqrt{6})(\sqrt{3})^{8 - 1} \)
= \((\sqrt{6})(27\sqrt{3}) = 27\sqrt{18} = 81\sqrt{2}\)