2019-II-36

Ans: C
Let $a$ and $r$ be the first term and the common ratio of the geometric sequence respectively.

$\{\begin{array}{ll}ar+a{r}^4=9& \dots \text{①}\\ a{r}^6+a{r}^9=288& \dots \text{②}\end{array}$

$\text{②}÷\text{①}$, we have

$\begin{array}{rcl}{\displaystyle \frac{a{r}^6+a{r}^9}{ar+a{r}^4}}& =& {\displaystyle \frac{288}{9}}\\ {\displaystyle \frac{a{r}^6(1+r^3)}{ar(1+r^3)}}& =& 32\\ r^5& =& 32\\ r& =& 2\end{array}$

Sub. $r=2$ into $\text{①}$, we have

$\begin{array}{rcl}a(2)+a(2{)}^4& =& 9\\ 2a+16a& =& 9\\ a& =& {\displaystyle \frac{1}{2}}\end{array}$

Therefore, the $20$th term of the sequence

$\begin{array}{cl}=& {\displaystyle \frac{1}{2}}(2{)}^{19}\\ =& 262144\end{array}$