Answer:
a) Total distance travelled to the nearest foot = 105ft
b) First 5 terms = 12, 9, 6.75, 5.0625, 3.796875
Step-by-step explanation:
From the question above, we can see that this is a geometric series or progression
Formula for geometric progression:
Tn = arⁿ-¹
Where
a = first term
r = common ration
n = number of terms
From the question, we are told that the ball dropped from 16 feet
Hence First term(a) = 16
We are told that the ball rebounds to 3/4 of it's previous height anytime it hits the surface.
Hence, common ratio(r) = 3/4
a) We are told to find total distance to the nearest foot that the ball travelled when it strikes the ground for the 10th time
This means we are to find the sum of the first 10 terms.
Because the ball rises and falls, we would use the rises form
Where the first term = 12 and the number of times is 9 because it rises 9 times
The formula we would use to find this sum is based on if the common ratio is less than or greater than 1
Our common ratio = 3/4, hence it is less than 1.
When r< 1, sum of n terms =
Sn = a(1 - rⁿ)/ 1 - r
It is important to
Where a = 12
r = 3/4
n = 9
Sn = 12( 1 - (3/4)^9) / 1- 3/4
Sn = 104.79ft
Approximately = 105ft
b) First five terms
Using the formula
Tn = arⁿ-¹
Where
a = first term = 12
r = common ration
n = number of terms
First term, n = 1 = 12 × 3/4^(1-1)
= 12 × 3/4 ^(0)
= 12× 1
= 12
Second term, n = 2 = 12 × 3/4^(2-1)
= 12 × 3/4 ^(1)
= 36/4
= 9
Third term, n = 3 = 12 × 3/4^(3-1)
= 12 × (3/4)²
= 6.75
Fourth term, n = 4= 12 × 3/4^(4-1)
= 12 × (3/4)³
= 5.0625
Fifth term, n = 5 = 12 × 3/4^(5-1)
= 12 × (3/4)⁴
= 3.796875
Therefore, the first 5 terms = 12, 9, 6.75, 5.0625, 3.796875
