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