Answer:
[tex]T_n = 7.143(0.7^n)[/tex]
Step-by-step explanation:
If a ball bounces to a height of 5feet, the initial drop = 5feet
If the ball rebounds to 70% of its previous height after each bounce, the next height after rebouncing will be expressed as;
= 70% of 5
= 70/100 * 5
= 0.7*5
= 3.5 feet
The next height will be 0.7*3.75 = 2.45 and so on
The heights wil form a sequence as thus;
5, 3.5, 2.45...
This sequence forms a geometric progression.
To get the function that could be used to model the nth term in the sequence of heights of the ball after the initial drop, w will find the nth term of the sequence
[tex]T_n = ar^{n-1}[/tex]
a is the first term of the sequence
n is the number of terms
r is the common ratio
From the sequence:
a = 5
[tex]r = \frac{3.5}{5} = \frac{2.75}{3.5} = 0.7[/tex]
Substitute the given values into the formula
[tex]T_n = 5(0.7)^{n-1}\\T_n = 5(\frac{0.7^n}{0.7^1} )\\T_n = \frac{5}{0.7}\times 0.7^n \\T_n = 7.143(0.7^n)[/tex]
The nth tern=m required is [tex]T_n = 7.143(0.7^n)[/tex]
