Chapter 10.3, Problem 57PPS

To determine

Find the distance it will raise in 5 minutes.

Answer to Problem 57PPS

After 5 minutes it will rise 193.75 ft.

Explanation of Solution

Given:

One minute after it is released, a gas − filled has risen 100 feet. In each succeeding minute, the balloon rises only 50% as far it rose in the previous minute. How far will it rise in 5 minutes?

Concept Used:

Let the height be h

Initial height: $h_1=100$

New height is previous height + rise amount (half of the previous rise amount):

  $\begin{array}{l}{h}_2=h_1+100(0.5)\\ $ h_3=h_2+100{(0.5)}^2$$ h_4=h_3+100{(0.5)}^3$\end{array}$

In summation form for t minutes: $h={\displaystyle \sum _{k=0}^{t-1}100·{(0.5)}^k}$

The general summation rule for GP series: ${\text{S}}_{\text{n}}=\frac{a_1(1-r^n)}{1-r}$

So, for our summation, we need the first $n=5$ terms.

Calculation:

In summation form for t minutes: $h={\displaystyle \sum _{k=0}^{t-1}100·{(0.5)}^k}$

The first term $a_1=100$ , the base of the exponential function is the common ratio $r=0.50$

  $a_1=100;r=0.50;n=5;\text{Find}{\text{S}}_{\text{n}}$

  $\begin{array}{l}{\text{S}}_{\text{n}}=\frac{a_1(1-r^n)}{1-r}\\ {\text{S}}_5=\frac{5·(1-{0.5}^5)}{1-0.5}=193.75\text{ft}\end{array}$

Thus, after 5 minutes it will rise 193.75 ft.