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