Answer
\[\frac{1}{2},\,\,\frac{1}{4},\,\,\frac{1}{8},\,\,\frac{1}{16},\,\,\frac{1}{32}\ \text{and}\ \frac{1}{64}\].
Work Step by Step
For the second term put \[n=2\] in the general formula stated above.
\[\begin{align}
& {{a}_{2}}={{a}_{1}}{{r}^{2-1}} \\
& =\frac{1}{2}\cdot {{\left( \frac{1}{2} \right)}^{1}} \\
& =\frac{1}{2}\cdot \frac{1}{2} \\
& =\frac{1}{4}
\end{align}\]
For the third term put \[n=3\] in the general formula stated above.
\[\begin{align}
& {{a}_{3}}={{a}_{1}}{{r}^{3-1}} \\
& =\frac{1}{2}\cdot {{\left( \frac{1}{2} \right)}^{2}} \\
& =\frac{1}{2}\cdot \frac{1}{4} \\
& =\frac{1}{8}
\end{align}\]
For the fourth term put \[n=4\] in the general formula stated above.
\[\begin{align}
& {{a}_{4}}={{a}_{1}}{{r}^{4-1}} \\
& =\frac{1}{2}\cdot {{\left( \frac{1}{2} \right)}^{3}} \\
& =\frac{1}{2}\cdot \frac{1}{8} \\
& =\frac{1}{16}
\end{align}\]
For the fifth term put \[n=5\] in the general formula stated above.
\[\begin{align}
& {{a}_{5}}={{a}_{1}}{{r}^{5-1}} \\
& =\frac{1}{2}\cdot {{\left( \frac{1}{2} \right)}^{4}} \\
& =\frac{1}{2}\cdot \frac{1}{16} \\
& =\frac{1}{32}
\end{align}\]
For the sixth term put \[n=6\] in the general formula stated above.
\[\begin{align}
& {{a}_{6}}={{a}_{1}}{{r}^{6-1}} \\
& =\frac{1}{2}\cdot {{\left( \frac{1}{2} \right)}^{5}} \\
& =\frac{1}{2}\cdot \frac{1}{32} \\
& =\frac{1}{64}
\end{align}\]
The first six terms of the geometric sequence are \[\frac{1}{2},\,\,\frac{1}{4},\,\,\frac{1}{8},\,\,\frac{1}{16},\,\,\frac{1}{32}\ \text{and}\ \frac{1}{64}\].
