The problem describes a scenario where a ball is dropped from a height of 15 feet, and each time it bounces, it reaches $\frac{4}{5}$ of its previous height. This setup is ideal for a geometric sequence since each bounce reaches a fraction of the previous height.

Let's go through each part:

Part A

To find a formula for $a_n$, where $a_n$ represents the height of the ball after the $n$-th bounce:

1. The initial height ($a_0$) is 15 feet.
2. Each subsequent height is $\frac{4}{5}$ of the previous height. So, this sequence has a common ratio $r = \frac{4}{5}$.

The formula for the $n$-th term of a geometric sequence is: $a_n = a_0 \cdot r^{n}$ Substituting the values we know: $a_n = 15 \cdot \left( \frac{4}{5} \right)^n$ This formula expresses $a_n$ as a function of $n$ for the height of the ball after $n$ bounces.

Part B

To find the height of the ball on the 6th bounce ($a_6$):

1. Substitute $n = 6$ into the formula from Part A: $a_6 = 15 \cdot \left( \frac{4}{5} \right)^6$

2. Calculate the value: $a_6 = 15 \cdot \left( \frac{4}{5} \right)^6 \approx 15 \cdot 0.262144 = 3.93216$ Rounding to three decimal places, the height on the 6th bounce is approximately: $a_6 \approx 3.932 \text{ feet}$

Summary of Answers

• Part A: $a_n = 15 \cdot \left( \frac{4}{5} \right)^n$
• Part B: The maximum height of the ball on the 6th bounce is approximately 3.932 feet.

Would you like further details or explanations on any part of this solution?

Here are some related questions to consider:

1. What would be the height of the ball on the 10th bounce?
2. How many bounces would it take for the ball to reach a height below 1 foot?
3. How does the formula change if the initial height were different, say 20 feet?
4. What happens if the common ratio changes to $\frac{3}{4}$ instead of $\frac{4}{5}$?
5. Could we generalize this formula for different initial heights and ratios?

Tip: In geometric sequences, the common ratio determines how quickly the terms decrease (if $|r| < 1$) or increase (if $|r| > 1$).