First part of question:
Find the general term that represents the situation in terms of k.
The general term for geometric series is:
[tex]a_{n}=a_{1}r^{n-1}[/tex]
[tex]a_{1}[/tex] = the first term of the series
[tex]r[/tex] = the geometric ratio
[tex]a_{1}[/tex] would represent the height at which the ball is first dropped. Therefore:
[tex]a_{1} = k[/tex]
We also know that the ball has a rebound ratio of 75%, meaning that the ball only bounces 75% of its original height every time it bounces. This appears to be our geometric ratio. Therefore:
[tex]r=\frac{3}{4}[/tex]
Our general term would be:
[tex]a_{n}=a_{1}r^{n-1}[/tex]
[tex]a_{n}=k(\frac{3}{4}) ^{n-1}[/tex]
Second part of question:
If the ball dropped from a height of 235ft, determine the highest height achieved by the ball after six bounces.
[tex]k[/tex] represents the initial height:
[tex]k = 235\ ft[/tex]
[tex]n[/tex] represents the number of times the ball bounces:
[tex]n = 6[/tex]
Plugging this back into our general term of the geometric series:
[tex]a_{n}=k(\frac{3}{4}) ^{n-1}[/tex]
[tex]a_{n}=235(\frac{3}{4}) ^{6-1}[/tex]
[tex]a_{n}=235(\frac{3}{4}) ^{5}[/tex]
[tex]a_{n}=55.8\ ft[/tex]
[tex]a_{n}[/tex] represents the highest height of the ball after 6 bounces.
Third part of question:
If the ball dropped from a height of 235ft, find the total distance traveled by the ball when it strikes the ground for the 12th time.
This would be easier to solve if we have a general term for the sum of a geometric series, which is:
[tex]S_{n}=\frac{a_{1}(1-r^{n})}{1-r}[/tex]
We already know these variables:
[tex]a_{1}= k = 235\ ft[/tex]
[tex]r=\frac{3}{4}[/tex]
[tex]n = 12[/tex]
Therefore:
[tex]S_{n}=\frac{(235)(1-\frac{3}{4} ^{12})}{1-\frac{3}{4} }[/tex]
[tex]S_{n}=\frac{(235)(1-\frac{3}{4} ^{12})}{\frac{1}{4} }[/tex]
[tex]S_{n}=(4)(235)(1-\frac{3}{4} ^{12})[/tex]
[tex]S_{n}=910.22\ ft[/tex]
