## Precalculus (6th Edition) Blitzer

$a_{n}=5(\dfrac{-1}{5})^{n-1}$ and $a_{7}=\dfrac{1}{3125}$
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 $a_{7}=\dfrac{1}{3125}$