Chapter ISG, Problem 7.6STP

To determine

To find: The geometric sequence that represents height of the ball, recursive formula of the sequence and the height of the ball on the 10^th bounce.

Answer to Problem 7.6STP

The required geometric sequence is $400,320256,204.8,\cdots$ , the recursive formula is $t_n=t_1\cdot {\left(0.8\right)}^{n-1}$ and the height of the ball at 10^th ounce is $53.7\text{ft}$ .

Explanation of Solution

Given information: The height from which a ball is dropped is 500 feet. The table that shows the first three bounces.

Bounce	1	2	3
Height (ft)	400	320	256

Formula used: The general formula of the geometric series is given by,

  $t_n=a\cdot r^{n-1}$

Here, $a$ is the first term and $r$ is the common ratio.

Calculation:

The common ratio of the given terms can be calculated as:

  $\begin{array}{l}\frac{320}{400}=0.8\\ \frac{256}{320}=0.8\end{array}$

So, the geometric sequence is $400,320256,204.8,\cdots$ .

The recursive formula for the sequence of height of balls is:

  $t_n=t_1\cdot {\left(0.8\right)}^{n-1}$

To find the height of the ball at 10^th ounce, substitute 10 for $n$ , 0.8 for $r$ , 400 for $t_1$ and 3 for $n$ in the above formula.

  $\begin{array}{c}{t}_{10}=400\cdot {\left(0.8\right)}^{10-1}\\ =400{\left(0.8\right)}^9\\ =53.69\text{ft}\end{array}$

The height of the ball to the nearest tenth is $53.7\text{ft}$ .

Therefore, the required geometric sequence is $400,320256,204.8,\cdots$ , the recursive formula is $t_n=t_1\cdot {\left(0.8\right)}^{n-1}$ and the height of the ball at 10^th ounce is $53.7\text{ft}$ .