Question

A recursive rule for a geometric sequence is a1=9; an=2/3 (an-1)

Answer 1

Since the first term is 9, then the second is

$\begin{gathered} a_2=9(\frac{2}{3}) \\ a_3=9(\frac{2}{3})^2 \\ \ldots \end{gathered}$

Then, ain iterative rule is

$a_n=9(\frac{2}{3})^{n-1}$

that is, the third one.