Question

Hello,Can you please help me with question# 5 in the picture? thank you

Answer 1

Solution

$a_1=9,for\text{ }n\ge2$$\begin{gathered} a_1=9 \\ a_2=\frac{2}{3a_{2-1}} \\ a_2=\frac{2}{3a_1} \\ a_2=\frac{2}{3(9)} \\ a_2=\frac{2}{27} \end{gathered}$$\begin{gathered} a_3=\frac{2}{3a_{3-1}} \\ a_3=\frac{2}{3a_2} \\ a_3=\frac{2}{3(\frac{2}{27})} \\ a_3=\frac{2}{\frac{2}{9}} \\ a_3=\frac{18}{2} \\ a_3=9 \end{gathered}$$\begin{gathered} a_4=\frac{2}{3a_3} \\ a_4=\frac{2}{3(9)} \\ a_4=\frac{2}{27} \end{gathered}$

Therefore for all values of n ≥ 2

$a_1=9,a_2=\frac{2}{27},a_3=9,a_4=\frac{2}{27}$