## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$(\dfrac{1}{10})^{10}=10^{-10}$
The $n^{th}$ term of a geometric sequence is given by the formula: $a_n = a_1 \ r^{(n-1)}(1)$ where, $a_1 = \ First \ Term$ and $r=\dfrac{a_2}{a_1}= \ Common \ Ratio$ We have: $a_1=1$ and $r=\dfrac{a_2}{a_1}=\dfrac{1/10}{1} =\dfrac{1}{10}$ Plug in $11$ for $n$ into equation (1) to obtain: $a_{11} = (1) \times (\dfrac{1}{10})^{(11-1)}=(\dfrac{1}{10})^{10}=10^{-10}$