Let $r$ be the common ratio of the geometric sequence. Since $a_7=32$ and $a_9=8$, then we have
$\left\{ \begin{array}{ll}
a_1r^6 = 32 & \ldots \unicode{x2460}\\
a_1r^8 = 8 & \ldots \unicode{x2461}
\end{array} \right.$
$\unicode{x2461} \div \unicode{x2460}$, we have
$\begin{array}{rcl}
r^2 & = & \dfrac{8}{32} \\
r & = & \pm \dfrac{1}{2}
\end{array}$
Therefore, we have
$\begin{array}{rcl}
a_1 (\pm \dfrac{1}{2})^6 & = & 32 \\
a_1 & = & 2048
\end{array}$
I is true. $a_1 = 2048 >0$.
II is true. For $r=\dfrac{1}{2}$, $a_1 = 2048$ and $a_2 = 1024$.
$\begin{array}{cl}
& a_1 – a_2 \\
= & 2048 – 1024 \\
= & 1024 \\
> & 0
\end{array}$
For $r=\dfrac{-1}{2}$, $a_1=2048$ and $a_2 = -1024$.
$\begin{array}{cl}
& a_1 – a_2 \\
= & 2048 – (-1024) \\
= & 3072 \\
> 0
\end{array}$
III may be false. For $r=\dfrac{-1}{2}$, $a_2 = -1024$.
$\begin{array}{cl}
& a_2 + a_3 + \cdots + a_{100} \\
= & \dfrac{(-1024)(1-(\frac{-1}{2})^{99})}{1-(\frac{-1}{2})} \\
< & 0
\end{array}$
