2009-II-43

Ans: A
Let $r$ be the common ratio of the geometric sequence. Since $a_7=32$ and $a_9=8$, then we have

$\{\begin{array}{ll}{a}_1{r}^6=32& \dots \text{①}\\ a_1{r}^8=8& \dots \text{②}\end{array}$

$\text{②}÷\text{①}$, we have

$\begin{array}{rcl}{r}^2& =& {\displaystyle \frac{8}{32}}\\ r& =& \pm {\displaystyle \frac{1}{2}}\end{array}$

Therefore, we have

$\begin{array}{rcl}{a}_1(\pm {\displaystyle \frac{1}{2}}{)}^6& =& 32\\ a_1& =& 2048\end{array}$

I is true. $a_1=2048>0$.

II is true. For $r={\displaystyle \frac{1}{2}}$, $a_1=2048$ and $a_2=1024$.

$\begin{array}{cl}& a_1–a_2\\ =& 2048–1024\\ =& 1024\\ >& 0\end{array}$

For $r={\displaystyle \frac{-1}{2}}$, $a_1=2048$ and $a_2=-1024$.

$\begin{array}{cl}& a_1–a_2\\ =& 2048–(-1024)\\ =& 3072\\ >0\end{array}$

III may be false. For $r={\displaystyle \frac{-1}{2}}$, $a_2=-1024$.

$\begin{array}{cl}& a_2+a_3+\cdots +a_{100}\\ =& {\displaystyle \frac{(-1024)(1-(\frac{-1}{2}{)}^{99})}{1-(\frac{-1}{2})}}\\ <& 0\end{array}$