I need some help with these quesitons

1. a1 = -1, and an = 4/3an-1 for n = 2, 3, 4, ___. What is the common ration, r, for this sequence?

2. A rubber ball is dropped from a height of 50 ft. Assume that each bounce of the ball is vertical and that each bounce reaches a height of 5/6 of the maximum height of the previous bounce. What is the total distance traveled throughout the lifetime of bounces for this ball?
I'm not sure I even understand this question

3. The following infinite geometric series will have a finite sum: 1001/6, 1001/36, 1001/216, 1001/1296
true?

1. since An = 4/3 An-1

An/An-1 = 4/3
That's r

#2. You should place it in the obvious context of geometric sequences
a = 50
r = 5/6
S = a/(1-r)

#3. What is r?
Divide any term by the previous one.
If |r| > 1, the infinite sum is finite.

So I'm getting false for #3...

For the 2nd one I'm getting 299..however my answer choices are 500, 450, 280, or the ball travels an infinite distance

The ball falls 50', and then starts bouncing.

Then it goes up and down on each 5/6 height bounce So that's a total of twice the Sum from then on. So, skipping the first term

S = 2*(50 * 5/6) /(1 - 5/6) = 500

This is almost a trick question. The immediate temptation, if the question is not read carefully, is just to say, Oh. S = 50/(1-5/6). But then they threw in the condition that you only count the bounces, and you have to recall that each complete bounce has an up and a down journey. I'm surprised that 600 was not one of the choices. It'd be an obvious but easy error to make.

#3. Eh? r = 1/6
each denominator is 6 times the previous one.

Did you do as I suggested?

r = A3/A2 = 1001/216 / 1001/36
= 1001/216 * 36/1001 = 1/6

Good luck on that SAT. You appear to have a long way to go.

Sure! I'll help you with each question step by step.

1. To find the common ratio, we need to look at the given sequence: a1 = -1 and an = (4/3) * an-1 for n = 2, 3, 4, ...

The common ratio (r) is the value that you multiply each term by to get the next term. In this case, we can see that each term is (4/3) times the previous term. So, the common ratio is 4/3.

2. This question is asking about the total distance traveled by a rubber ball when it is dropped from a height of 50 ft, assuming each bounce goes up to a height of (5/6) times the maximum height of the previous bounce.

To calculate the total distance traveled, we need to consider two types of distances:
- The distances traveled during the ascents (going up to the maximum height)
- The distances traveled during the descents (coming back down to the ground)

Let's break it down step by step:
- The ball is dropped from a height of 50 ft, which is the first descent. This descent distance is 50 ft.
- The first bounce goes up to (5/6) * 50 ft, and then comes back down to the ground, so the distance for this bounce is (5/6) * 50 + 50 ft.
- The second bounce goes up to (5/6) * ((5/6) * 50) ft, and then comes back down to the ground, so the distance for this bounce is (5/6)^2 * 50 + (5/6) * 50 + 50 ft.
- The third bounce goes up to (5/6) * ((5/6)^2 * 50) ft, and then comes back down to the ground, so the distance for this bounce is (5/6)^3 * 50 + (5/6)^2 * 50 + (5/6) * 50 + 50 ft.

To find the total distance, we need to sum up all these distances over an infinite number of bounces. However, since each bounce is getting diminishingly smaller, we can use the formula for the sum of an infinite geometric series:

Sum = a / (1 - r)

In this case:
- a is the first term, which is 50 ft.
- r is the common ratio, which is (5/6).
So, the total distance traveled by the ball throughout its lifetime of bounces is given by:
Sum = 50 / (1 - (5/6))

3. The question is asking whether the given infinite geometric series will have a finite sum: 1001/6, 1001/36, 1001/216, 1001/1296.

To determine if the series has a finite sum, we need to check if the absolute value of the common ratio is less than 1. If it is, then the series will have a finite sum; otherwise, it will diverge to infinity.

Let's calculate the common ratio:
First term, a1 = 1001/6
Second term, a2 = 1001/36

The common ratio (r) can be found by dividing any term by the previous term in the series. So:
r = a2 / a1 = (1001/36) / (1001/6) = (1001/36) * (6/1001) = 1/6

Since the absolute value of the common ratio (1/6) is less than 1, the series will have a finite sum. Therefore, the statement is true.

I hope this helps! Let me know if you have any further questions.