Answer
$1, \frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \frac{1}{81}$
Work Step by Step
The general term $a_n$ of a geometric sequence is given by $a_n = a_1r^{n-1}$ where $a_1$ is the first term and $r$ is the common ratio.
We plug in $n = 1, 2, 3, 4, 5$ to find the first five terms of the geometric sequence
$a_1 = 1\cdot\frac{1}{3}^{1-1} = 1\cdot\frac{1}{3}^0=1\cdot1=1$
$a_2 = 1\cdot\frac{1}{3}^{2-1} = 1\cdot\frac{1}{3}^1=1\cdot\frac{1}{3}=\frac{1}{3}$
$a_3 = 1\cdot\frac{1}{3}^{3-1} = 1\cdot\frac{1}{3}^2=1\cdot\frac{1}{9}=\frac{1}{9}$
$a_4 = 1\cdot\frac{1}{3}^{4-1} = 1\cdot\frac{1}{3}^3=1\cdot\frac{1}{27}=\frac{1}{27}$
$a_5 = 1\cdot\frac{1}{3}^{5-1} = 1\cdot\frac{1}{3}^4=1\cdot\frac{1}{81}=\frac{1}{81}$
