Answer
$$r = 2{\text{ and }}{a_n} = 6{\left( 2 \right)^{n - 1}}$$
Work Step by Step
$$\eqalign{
& {\text{the terms of the sequence are }} \cr
& {a_1} = 6,{a_2} = 12,{a_3} = 24,{a_4} = 48,... \cr
& {\text{to verify that the sequence is geometric}}{\text{, divide each term except the first by }} \cr
& {\text{the preceding term}} \cr
& \frac{{48}}{{24}} = \frac{{24}}{{12}} = \frac{{12}}{6} = 2 \cr
& {\text{the ratio is constant}}{\text{, so the sequence is geometric with }}r = 2 \cr
& {\text{The general term of a geometric sequence is }}{a_n} = {a_1}{r^{n - 1}} \cr
& {\text{then}} \cr
& {a_n} = 6{\left( 2 \right)^{n - 1}} \cr} $$
