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