Ans: C
Let $a$ and $r$ be the first term and the common ratio of the geometric sequence respectively.
Let $a$ and $r$ be the first term and the common ratio of the geometric sequence respectively.
$\left\{ \begin{array}{ll}
ar + ar^4 = 9 & \ldots \unicode{x2460} \\
ar^6 + ar^9 = 288 & \ldots \unicode{x2461}
\end{array} \right.$
$\unicode{x2461} \div \unicode{x2460}$, we have
$\begin{array}{rcl}
\dfrac{ar^6 + ar^9}{ar + ar^4} & = & \dfrac{288}{9} \\
\dfrac{ar^6(1 + r^3)}{ar(1 + r^3)} & = & 32 \\
r^5 & = & 32 \\
r & = & 2
\end{array}$
Sub. $r = 2$ into $\unicode{x2460}$, we have
$\begin{array}{rcl}
a(2) + a(2)^4 & = & 9 \\
2a + 16 a & = & 9 \\
a & = & \dfrac{1}{2}
\end{array}$
Therefore, the $20$th term of the sequence
$\begin{array}{cl}
= & \dfrac{1}{2} (2)^{19} \\
= & 262\ 144
\end{array}$
