Answer
$\frac{5}{1024}\text{ ft}$.
Work Step by Step
1. Familiarize yourself with the problem.
The rebound distance from a geometric sequence is:
$\frac{1}{4}\cdot 20\text{ ft},{{\left( \frac{1}{4} \right)}^{2}}\cdot 20\text{ ft},{{\left( \frac{1}{4} \right)}^{3}}\cdot 20\text{ ft},\ldots =5\text{ ft},\left( \frac{1}{4} \right)\cdot 5\text{ ft},{{\left( \frac{1}{4} \right)}^{2}}\cdot 5\text{ ft},\ldots $
The height of the ${{6}^{\text{th}}}$ rebound is the 6th term of the sequence.
2. Translate the problem into an equation.
The above sequence is a geometric sequence for which,
${{a}_{1}}=5\text{ ft}$
And,
$\begin{align}
& r=\frac{\frac{1}{4}\cdot 5\text{ ft}}{5\text{ ft}} \\
& =\frac{1}{4}
\end{align}$
And,
$n=6$
${{a}_{n}}={{a}_{1}}{{r}^{n-1}}$,
So,
$\begin{align}
& {{a}_{n}}={{a}_{1}}{{r}^{n-1}} \\
& {{a}_{6}}=\left( 5\text{ ft} \right){{\left( \frac{1}{4} \right)}^{6-1}}
\end{align}$
3. Carry out the mathematical operations to solve the equation.
Substitute $n=6, {{a}_{1}}=5 \text{ ft}$ and $r=\frac{1}{4}$,
$\begin{align}
& {{a}_{6}}=\left( 5\text{ ft} \right){{\left( \frac{1}{4} \right)}^{6-1}} \\
& =\left( 5\text{ ft} \right){{\left( \frac{1}{4} \right)}^{5}} \\
& =\frac{5}{1024}\text{ ft}
\end{align}$
Thus, the rebound height is$\frac{5}{1024}\text{ ft}$ the 6th time
