Answer
$ a_{n}=5(\dfrac{-1}{5})^{n-1}$ and $ a_{7}=\dfrac{1}{3125}$
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{-1}{5}$ and $ n=7$
Now, $ a_{n}=5(\dfrac{-1}{5})^{n-1}$
Now, $ a_{7}=5(\dfrac{-1}{5})^{7-1}=5(\dfrac{-1}{5})^{6}=\dfrac{1}{3125}$
Our answers are: $ a_{n}=5(\dfrac{-1}{5})^{n-1}$ and $ a_{7}=\dfrac{1}{3125}$
