The $n$th term of a sequence is given.
\[
a_{n}=7(3)^{n-1}
\]
(a) Find the first five terms of the sequence.
\[
\begin{array}{l}
a_{1}= \\
a_{2}= \\
a_{3}= \\
a_{4}= \\
a_{5}=
\end{array}
\]
(b) What is the common ratio $r$ ?
\[
r=
\]

Step 1 ：Given the $n$th term of a sequence is $a_{n}=7(3)^{n-1}$.

Step 2 ：Substitute $n=1,2,3,4,5$ into the formula $a_{n}=7(3)^{n-1}$ to find the first five terms of the sequence.

Step 3 ：The first five terms of the sequence are $a_{1}=7$, $a_{2}=21$, $a_{3}=63$, $a_{4}=189$, and $a_{5}=567$.

Step 4 ：The common ratio $r$ of a geometric sequence is the ratio between any term and its preceding term. In this case, it is $3$ because each term is multiplied by $3$ to get the next term.

Step 5 ：So, the common ratio $r$ is $3$.

Step 6 ：Final Answer: \[\begin{array}{l} a_{1}= \boxed{7} \\ a_{2}= \boxed{21} \\ a_{3}= \boxed{63} \\ a_{4}= \boxed{189} \\ a_{5}= \boxed{567} \\ r= \boxed{3} \end{array}\]