Progressions and Sequences

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Progressions and Sequences

by surabhibahl » Sun Sep 15, 2013 12:45 am
The question is as follows:

Height, to which a ball reaches, reduces to the same ratio after each bounce. It is dropped from a height of 100 feet and after 2 bounces it reaches to a height of 25 feet. What is the factor to which height is reduced after each bounce?

The correct answer is 2 and a half.

Please tell me the fastest approach and the formula to be used?

Thanks
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by Brent@GMATPrepNow » Sun Sep 15, 2013 8:05 am
surabhibahl wrote:The question is as follows:

Height, to which a ball reaches, reduces to the same ratio after each bounce. It is dropped from a height of 100 feet and after 2 bounces it reaches to a height of 25 feet. What is the factor to which height is reduced after each bounce?

The correct answer is 2 and a half.

Please tell me the fastest approach and the formula to be used?

Thanks
Hmmm, this question has some interesting phrasing . . . "reduces to the same ratio?"

If the height is 25 feet after 2 bounces it seems that the height is reduced by 1/2 after each bounce.

Start at 100 feet
1 bounce: rises to 50 feet
2 bounces: rises to 25 feet
3 bounces: rises to 12.5 feet
4 bounces: rises to 6.25 feet

I'm not sure what you mean when you say that the correct answer is 2 and a half.
What's the source of this question?

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by surabhibahl » Sun Sep 15, 2013 11:08 am
Hi Brent,

I found this in a GMAT help book.

Following is the answer, can you explain the solution then?

Heights after each bounce are in GP. Initial height is 100k.
So, height after 1st bounce is 100k
Height after 2nd bounce is 100k^2 ( where k is the ratio to which height is reduced )
100k^2 = 25

k= 1/2 (Sorry not 2 and a half)

Can you explain the formula and what the logic behind this is?

Thanks a ton!!

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by Alpha114 » Sun Sep 15, 2013 11:33 am
The moment i saw this Q, compound interest formula came to my mind,

25= 100(1-x)^2.

Solved for x . x=1/2.

:)

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by surabhibahl » Tue Sep 17, 2013 8:48 am
Can you tell me how to solve this using GP?

Thanks!

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by surabhibahl » Tue Sep 17, 2013 8:48 am
Can you tell me how to solve this using GP?

Thanks!

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by Brent@GMATPrepNow » Tue Sep 17, 2013 8:55 am
surabhibahl wrote:Can you tell me how to solve this using GP?

Thanks!
Sure.
With each bounce, the new height is some fraction of the previous height.
Let's let k = this fraction

Starting height: 100 feet
1 bounce: new height = 100(k)
2 bounces: new height = 100(k)(k)
3 bounces: new height = 100(k)(k)(k)
4 bounces: new height = 100(k)(k)(k)(k)
etc.

If, after 2 bounces, the height is 25 feet, we can write: 100(k)(k) = 25
In other words, 100k² = 25
Divide both sides by 100 to get: k² = 1/4
So, k = 1/2

Does that help, or are you wanting to plug values into some formula?

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by surabhibahl » Tue Sep 17, 2013 11:02 am
Could you probably plug it in a formula and explain ?

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by StrawberryCow » Tue Oct 29, 2013 11:29 pm
Hi Brent,
How many questions can be expected from Progressions and Sequences?
Apart from AP,GP & HP what are the other types of questions that we can expect from this topic.

Best Regards

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by Brent@GMATPrepNow » Wed Oct 30, 2013 7:40 am
surabhibahl wrote:Can you tell me how to solve this using GP?
Thanks!
I should note that I'm not a big fan of memorizing formulas, since formulas have the potential for silly mistakes. For this question, I'd rather list some values and apply some logic.

Having said that, I do know that many students relish in memorizing formulas, so . . .

Some background: a geometric progression (GP) is a sequence in which each term is obtained by multiplying the previous term by some constant, r (called the constant ratio)
Here are some examples of geometric progressions"
2, 6, 18, 54, 162, ... (here, r = 3)
160, 80, 40, 20, 10, ... (here, r = 1/2)
7, -14, 28, -56, 112, ... (here, r = -2)

In a geometric progression, term n = ar^(n-1), where a = term1

So, how do we use the geometric progression formula in the following question?
Height, to which a ball reaches, reduces to the same ratio after each bounce. It is dropped from a height of 100 feet and after 2 bounces it reaches to a height of 25 feet. What is the factor to which height is reduced after each bounce?
This question illustrates the potential danger of memorizing formulas.

It is dropped from a height of 100 feet
The first term (a) in this sequence is 100 (no problem)

. . . after 2 bounces it reaches to a height of 25 feet
So, does term2 = 25?
If we assume that this is the case, we'll answer the question incorrectly. Here's why:
Term1 (which is 100) represents the height after zero bounces
So, term2 represents the height after 1 bounce
And term3 represents the height after 2 bounces
In other words, term3 = 25

Since term n = ar^(n-1), we can see that term3 = ar^(3-1) = ar²
Since a = 100 and term3 = 25, we get: 25 = 100r²
Solve for r to get r = 1/2 or -1/2
Given the real-world context of this question, we can conclude that r = 1/2
So, each term is determined by multiplying the previous term by 1/2

Our sequence looks like this: 100, 50, 25, 12.5, 6.25, . . .

Cheers,
Brent
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by Brent@GMATPrepNow » Wed Oct 30, 2013 7:48 am
StrawberryCow wrote:Hi Brent,
How many questions can be expected from Progressions and Sequences?
Apart from AP,GP & HP what are the other types of questions that we can expect from this topic.

Best Regards
I think a student might see one question involving sequences.
NOTE: the GMAT does not require you to know the terms arithmetic progression (AP), geometric progression (GP) and harmonic progression (HP). Instead, the GMAT will define the sequence for you.
Also, you aren't expected to know the formulas for arithmetic, harmonic or geometric sequences or series.

To familiarize yourself with these kinds of questions, you can use BTG's tagging feature. Here are all of the questions tagged as Sequences and Series questions: https://www.beatthegmat.com/forums/tags/ ... and-series


Cheers,
Brent
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