Geometric Series

A

aishasoleja

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A bouncing balls reaches heights of 16 cm, 12.8 cm and 10.24 cm on three consecutive bounces.

a) If the ball started at a height of 25 cm, how many times has it bounced when it reaches a height of 16 cm?
b) Write a geometric series for the downward distances the ball travels from its release at 25 cm.
c) Write a geometric series for the upward distances the ball travels from its first bounce.
d) Find the total vertical distance the ball travels before it comes to rest.

Ok here is what I get

a) no clue

b) no clue

c) no clue

d) no clue

I am having trouble with this question and I would appreciate the help THANK U
 
aishasoleja said:
A bouncing balls reaches heights of 16 cm, 12.8 cm and 10.24 cm on three consecutive bounces.

a) If the ball started at a height of 25 cm, how many times has it bounced when it reaches a height of 16 cm?
b) Write a geometric series for the downward distances the ball travels from its release at 25 cm.
c) Write a geometric series for the upward distances the ball travels from its first bounce.
d) Find the total vertical distance the ball travels before it comes to rest.

Ok here is what I get

a) no clue

b) no clue

c) no clue

d) no clue

I am having trouble with this question and I would appreciate the help THANK U
Click to expand...

I'll do a different problem to give you some clue:

A bouncing balls reaches heights of 16 cm, 8 cm and 4 cm on three consecutive bounces.

a) If the ball started at a height of 128 cm, how many times has it bounced when it reaches a height of 16 cm?
b) Write a geometric series for the downward distances the ball travels from its release at 128 cm.
c) Write a geometric series for the upward distances the ball travels from its first bounce.
d) Find the total vertical distance the ball travels before it comes to rest.

So our geometric series has ratio

\(\displaystyle r \, = \, \frac{a_2}{a_1} \, = \, \frac{8}{16} \, = \frac{1}{2} \, = \, 0.5\)

So our new series starts with new start number

\(\displaystyle a_1 \, = \, 128\)

with same ratio of reduction.

we know for a geometric series

a(n) = a(1) * r^(n-1)

for problem

a)

16 = 128 * (0.5)^(n-1)

1/8 = (0.5)^(n-1)

taking log and simplifying

n = 4

So on 3 rd bounce it will reach the height of 16 cm.

b) this geometric series starts with 128 - with a reduction ratio 1/2. So the series is

128, 64, 32, 16,.....

c) This geometric series starts with 64 (the height acheived after first bounce) - with a reduction ratio 1/2. So the series is

64, 32, 16,.....

Now you should have some clues to do your work.

Please show us your work, indicating exactly where you are stuck, so that we know where to begin to help you.
 
A bouncing balls reaches heights of 16 cm, 12.8 cm and 10.24 cm on three consecutive bounces.

a) If the ball started at a height of 25 cm, how many times has it bounced when it reaches a height of 16 cm?
b) Write a geometric series for the downward distances the ball travels from its release at 25 cm.
c) Write a geometric series for the upward distances the ball travels from its first bounce.
d) Find the total vertical distance the ball travels before it comes to rest.
Click to expand...

Hello, aishasoleja,

To add to what Subhotosh has given you (since he was too quick for me :) :

The first step in a geometric sequence or series problem is to find the “common ratio,” r. Take any two consecutive terms and divide the “later” term by the “earlier” term. For example in the sequence 32, 16, 8, 4, 2…, one can find r as follows:

r = 16/32 = 2
r = 8/16 = 2
r = 4/8 = 2

The r value must be the same everywhere in the sequence.

Now suppose the above sequence describes the bounces of a ball. The downward travel sequence is 32, 16, 8, 4, 2… (because the ball was dropped from a height of 32 initially), and the upward bounce sequence is 16, 8, 4, 2…, because 16 was the height of the first bounce. Make sense?

The total vertical travel of the ball is the sum of the numbers in the two sequences. When the terms are added together, it is called a geometric series (not a sequence):

S1 = 32 + 16 + 8 + 4 + …
S2 = 16 + 8 + 4 + 2 +…

One can find the sum of each series separately and add them together. Here are some links with info:

http://en.wikipedia.org/wiki/Geometric_progression

http://en.wikipedia.org/wiki/Geometric_series
 
Thank you both for your help ! Subhotosh and wjm11 :) Thank you for you detailed reply. God Bless you :D

Okay Here is my try:

A bouncing balls reaches heights of 16 cm, 12.8 cm and 10.24 cm on three consecutive bounces.

a) If the ball started at a height of 25 cm, how many times has it bounced when it reaches a height of 16 cm?
b) Write a geometric series for the downward distances the ball travels from its release at 25 cm.
c) Write a geometric series for the upward distances the ball travels from its first bounce.
d) Find the total vertical distance the ball travels before it comes to rest.


R= a2/a1= 12.8 \ 16 = .8 >>>> common ratio

We have two series
One that starts with 25
One that starts with 16
Both have the same reduction as you said
Okay now with the answers

a) an = a1 * r^n-1
So 16=25*(.8)^n-1
16\25 = (.8)^n-1
Okay I was stuck here because I don't get what u mean by taking the log


b)this geometric series starts with 25, 25(.8), …

So it's: 25, 20, 16, 12.8, 10.24, 8.19, 6.55, 5.24, 4.19, 3.35, 2.68, 2.15, 1.72, 1.37, 1, .88, .70, .56, .45, .36, .29, .23, .18, .14, .11, 0.09, 0.07, .0.06, .05, .04, .03, .02, .01

Is this right, or is it an infinite series? I got stuck here because I don't know for sure that it's infinite

c)20, 16, 12.8, 10.24, 8.19, 6.55, 5.24, 4.19, 3.35, 2.68, 2.15, 1.72, 1.37, 1, .88, .70, .56, .45, .36, .29, .23, .18, .14, .11, 0.09, 0.07, .0.06, .05, .04, .03, .02, .01

Again I am not sure if it should be infinite or finite, but my guess is that it's finite because it can't be negative since we're talking about heights, right?

d)The total vertical travel of the ball is the sum of the numbers in the two sequences.
So does this mean I just add what I found in b and c to get the answer?
Or could I use:
Sn= a1(1-r^n) / (1-r) >>>>>>>> sum of a finite geometric series ??? Yes or no
If Yes how do I find n accurately?
 
aishasoleja said:
Thank you both for your help ! Subhotosh and wjm11 :) Thank you for you detailed reply. God Bless you :D

Okay Here is my try: Good work

A bouncing balls reaches heights of 16 cm, 12.8 cm and 10.24 cm on three consecutive bounces.

a) If the ball started at a height of 25 cm, how many times has it bounced when it reaches a height of 16 cm?
b) Write a geometric series for the downward distances the ball travels from its release at 25 cm.
c) Write a geometric series for the upward distances the ball travels from its first bounce.
d) Find the total vertical distance the ball travels before it comes to rest.


R= a2/a1= 12.8 \ 16 = .8 >>>> common ratio

We have two series
One that starts with 25
One that starts with 16
Both have the same reduction as you said
Okay now with the answers

a) an = a1 * r[sup:3p5ztnc8](n-1)[/sup:3p5ztnc8]
So 16=25*(.8)[sup:3p5ztnc8]n-1[/sup:3p5ztnc8]
16\25 = (.8)[sup:3p5ztnc8]n-1[/sup:3p5ztnc8]

(0.8)[sup:3p5ztnc8]2[/sup:3p5ztnc8] = (.8)[sup:3p5ztnc8]n-1[/sup:3p5ztnc8]

n = 3

Solution by taking log

log(16\25) = log((.8)[sup:3p5ztnc8]n-1[/sup:3p5ztnc8])

log(.64) = (n-1) * log(0.8)

n-1 = log (0.64)/log(0.8) = -0.193820026/-0.096910013 = 2

Okay I was stuck here because I don't get what u mean by taking the log


b)this geometric series starts with 25, 25(.8), …

So it's: 25, 20, 16, 12.8, 10.24, 8.19, 6.55, 5.24, 4.19, 3.35, 2.68, 2.15, 1.72, 1.37, 1, .88, .70, .56, .45, .36, .29, .23, .18, .14, .11, 0.09, 0.07, .0.06, .05, .04, .03, .02, .01 <<< This is a sequence not a series - to get series you need to add these up

Is this right, or is it an infinite series? I got stuck here because I don't know for sure that it's infinite - what is your question asking you?
c)20, 16, 12.8, 10.24, 8.19, 6.55, 5.24, 4.19, 3.35, 2.68, 2.15, 1.72, 1.37, 1, .88, .70, .56, .45, .36, .29, .23, .18, .14, .11, 0.09, 0.07, .0.06, .05, .04, .03, .02, .01

Again I am not sure if it should be infinite or finite, but my guess is that it's finite because it can't be negative since we're talking about heights, right?

d)The total vertical travel of the ball is the sum of the numbers in the two sequences.
So does this mean I just add what I found in b and c to get the answer?
Or could I use:
Sn= a1(1-r^n) / (1-r) >>>>>>>> sum of a finite geometric series ??? Yes or no --Yes and no

when r < 1 and n becomes very very large (like infinity) - what happens to the value of r[sup:3p5ztnc8]n[/sup:3p5ztnc8]

Yes you'll use that equation - but you need to use it as n becomes very very very large.If Yes how do I find n accurately?
Click to expand...
 
Okay So I understood a :D

b) after adding them up I got 124.75

c) after adding them up I got 99.75

d) would be 99.75+124.75=224.5


When I calculated Sn I got 124.9 (so it's close)

When I calculated Sn for the other one I got (99.9) >>> again very close


I feel so proud of myself :) Thank you again for your generous help :D :D You really are a math expert ! :)
 
aishasoleja said:
Okay So I understood a :D

b) after adding them up I got 124.75

Should be expressed as

S[sub:jhulmcmf]n[/sub:jhulmcmf] = 25 + 20+ 16+ ... + 25*(0.8)[sup:jhulmcmf]n-1[/sup:jhulmcmf]

c) after adding them up I got 99.75

Should be expressed as

S[sub:jhulmcmf]n[/sub:jhulmcmf] = 20 +16 + 12.8 + ... + 20*(0.8)[sup:jhulmcmf]n-1[/sup:jhulmcmf]


d) would be 99.75+124.75=224.5.... Not quite...

When n becomes very large, (0.8)[sup:jhulmcmf]n[/sup:jhulmcmf] = 0

so

S[sub:jhulmcmf]inf[/sub:jhulmcmf] = 25/(1-0.8) + 20/(1-0.8) = 125 + 100 = 225

When I calculated Sn I got 124.9 (so it's close)

When I calculated Sn for the other one I got (99.9) >>> again very close


I feel so proud of myself :) Thank you again for your generous help :D :D You really are a math expert ! :)
Click to expand...
 
I don't get what you mean by:
When n becomes very large, (0.8)n = 0 >>> when n becomes very large then 0.8^n-1 would be less than zero yes but we're going to mulitply it with a1 and that will solve our problem

Let me explain my point of view:
n was approximatly 32 for the series: 25+20+16+...
n was approcimatly 31 for the series: 20+16+...

ur right it does give a big number, but when u multiply it with a1 it gives u a reasonable number

For example: (Let's take the first one)
Sn= 25(1-r^n-1)\1-r >>>>> 25(1-.8^32)\1-.8>>>>>> 25-.02\1-.8=125
The second was about 100

So total is 125+100=225 (like ur answer)

Why is this wrong?
 
aishasoleja said:
I don't get what you mean by:

When n becomes very large, (0.8)n = 0 …
Click to expand...


Is your confusion a result of reading it as you typed it?

0.8 is not being multiplied by n; 0.8 is being raised to the power of n.

Get a calculator.

Start multiplying 0.8 by itself, repeatedly.

You'll see that the products are moving toward zero.

 
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