Answer
$ a_{n}=18(\dfrac{1}{3})^{n-1}$ and $ a_{7}=\dfrac{2}{81}$
Work Step by Step
The general formula to find the nth term of a Geometric sequence is given as: $ a_{n}=a_1r^{n-1}$
$ r=\dfrac{a_2}{a_1}=\dfrac{6}{18}=\dfrac{1}{3}$ and $ n=7$
Now, $ a_{n}=18(\dfrac{1}{3})^{n-1}$
Now, $ a_{7}=18(\dfrac{1}{3})^{7-1}=18 \times (\dfrac{1}{3})^{6}=\dfrac{2}{81}$
Our answers are: $ a_{n}=18(\dfrac{1}{3})^{n-1}$ and $ a_{7}=\dfrac{2}{81}$
