Answer
The sequence $12,6,3$ repeats indefinitely
Work Step by Step
Consider $a_1=7$ and calculate the first ten terms:
$$\begin{align*}
a_1&=7\\
a_2&=3(7)+3=24\\
a_3&=\dfrac{24}{2}=12\\
a_4&=\dfrac{12}{2}=6\\
a_5&=\dfrac{6}{2}=3\\
a_6&=3(3)+3=12\\
a_7&=\dfrac{12}{2}=6\\
a_8&=\dfrac{6}{2}=3\\
a_9&=3(3)+3=12\\
a_{10}&=\dfrac{12}{2}=6.
\end{align*}$$
We notice that the sequence $12,6,3$ repeats indefinitely.
Consider $a_1=8$ and calculate the first ten terms:
$$\begin{align*}
a_1&=8\\
a_2&=\dfrac{8}{2}=4\\
a_3&=\dfrac{4}{2}=2\\
a_4&=\dfrac{2}{2}=1\\
a_5&=3(1)+3=6\\
a_6&=\dfrac{6}{2}=3\\
a_7&=3(3)+3=12\\
a_8&=\dfrac{12}{2}=6\\
a_9&=\dfrac{6}{2}=3\\
a_{10}&=3(3)+3=12.
\end{align*}$$
We notice that the sequence $12,6,3$ repeats indefinitely.
Consider $a_1=9$ and calculate the first ten terms:
$$\begin{align*}
a_1&=9\\
a_2&=3(9)+3=30\\
a_3&=\dfrac{30}{2}=15\\
a_4&=3(15)+3=48\\
a_5&=\dfrac{48}{2}=24\\
a_6&=\dfrac{24}{2}=12\\
a_7&=\dfrac{12}{2}=6\\
a_8&=\dfrac{6}{2}=3\\
a_9&=3(3)+3=12\\
a_{10}&=\dfrac{12}{2}=6.
\end{align*}$$
We notice that the sequence $12,6,3$ repeats indefinitely.
