SOLUTION: 3. A ball is dropped vertically from a height of 50 meters. After each bounce, the ball reaches a maximum height equal to 80 percent of the maximum height of the previous bounce.

Question 867322: 3. A ball is dropped vertically from a height of 50 meters. After each bounce, the ball reaches a maximum height equal to 80 percent of the maximum height of the previous bounce.
a.) Write the first three terms of the sequence. Explain how you know what they are.
b.) Find the height to the nearest tenth of a meter of the ball after the tenth bounce. Show and explain your work.
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
drops from 50 m
height reached = 0.8 time previous height
h2= 0.8*50 = 40
h3=0.8*40=32
The first three terms are 50, 40, 32
t2/t1= 50/40 = 5/4
t3/t2= 40/32 = 5/4
The ratio of the term and the preceeding term is a constant. The common ratio. and less than 1
It is a geometric sequence
after 10 bounces

tn=a*r^(n-1)
nth. term of a GP
a=50
r=0.8
n=10
t10 = 50*0.8^9
=6.71 m
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