#### Answer

The first five terms are:
$a_1 =6
\\a_2 =18
\\a_3 =54
\\a_4= 162
\\a_5=486$
The sequence is geometric with a common ratio of $3$.
The $n^{th}$ term of the geometric sequence is: $a_n = 6\cdot(3^{n-1})$

#### Work Step by Step

To find the first five terms, substitute 1, 2, 3, 4, and 5 to the given formula to obtain:
$a_1 = 2(3^1) = 2(3) = 6
\\a_2 = 2(3^2) = 2(9) = 18
\\a_3 = 2(3^3) = 2(27) = 54
\\a_4=2(3^4) = 2(81) = 162
\\a_5=2(3^5) = 2(243) = 486$
RECALL:
A sequence is geometric if there is a common ratio among consecutive terms.
Notice that 3 times a term is equal to the value of the next term.
This means that the sequence has a common ratio of $3$.
Thus, the sequence is geometric with a common ratio of $3$.
The $n^{th}$ term of a geometric sequence is given by the formula $a_n = ar^{n-1}$
where $a$ = first term and $r$= common ratio.
This means that the $n^{th}$ term of the given geometric sequence, whose $a=6$ and $r=3$, is:
$a_n = 6\cdot(3^{n-1})$