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 Post subject: P1, P2, P3, …. Pn, Pn+1 = Pn/3
 Post Posted: Tue May 29, 2012 3:37 am 
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Students


Posts: 12
Q) A series of positive numbers P1, P2, P3, …. Pn, Pn+1 = Pn/3 for all values of n ≥ 1. what is the value of 2P4?
(1) P2 = 1/3
(2) P2 - P3 = 2/9

OA is D
need some help in going about this type of problem.


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 Post subject: Re: P1, P2, P3, …. Pn, Pn+1 = Pn/3
 Post Posted: Sat Jun 02, 2012 10:54 pm 
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ManhattanGMAT Staff


Posts: 2390
Hey PudGe, again, according to forum guidelines, we need an original source for each question. Please provide and we're happy to help. :-)

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Jamie Nelson
ManhattanGMAT Instructor


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 Post subject: Re: P1, P2, P3, …. Pn, Pn+1 = Pn/3
 Post Posted: Tue Jun 26, 2012 2:29 pm 
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Students


Posts: 1
Hello Nelson,

This question is from the Mock test of Jamboree classes in India. This is from their free test, so It shouldn't be an issue to solve it.

This is a challenging question though...:p

Thanks,


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 Post subject: Re: P1, P2, P3, …. Pn, Pn+1 = Pn/3
 Post Posted: Wed Jun 27, 2012 9:32 am 
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Forum Guests


Posts: 125
Hi,

I believe the OA is D. Please correct me if I am wrong.


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 Post subject: Re: P1, P2, P3, …. Pn, Pn+1 = Pn/3
 Post Posted: Fri Jun 29, 2012 6:42 pm 
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Forum Guests


Posts: 125
Hi,

The question does mention that Pn+1 = Pn/3 which is the equivalent of your statement. So I am unable to find anything wrong with the question. Could you please share what is wrong with the question. We do need the statements (each statement would suffice by itself) to provide us the first value of the expression.


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 Post subject: Re: P1, P2, P3, …. Pn, Pn+1 = Pn/3
 Post Posted: Sun Jul 01, 2012 10:57 pm 
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ManhattanGMAT Staff


Posts: 2390
krishnan.anju1987 wrote:
Hi,

The question does mention that Pn+1 = Pn/3 which is the equivalent of your statement. So I am unable to find anything wrong with the question. Could you please share what is wrong with the question. We do need the statements (each statement would suffice by itself) to provide us the first value of the expression.


Thanks krishnan! I get it now . . . any value in this sequence is the previous value divided by 3. This is an unusual way to express this and that is what threw me. Usually the testwriters would express this as Pn= P(n-1)/3.

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Jamie Nelson
ManhattanGMAT Instructor


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 Post subject: Re: P1, P2, P3, …. Pn, Pn+1 = Pn/3
 Post Posted: Sun Jul 01, 2012 11:06 pm 
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ManhattanGMAT Staff


Posts: 2390
PudGe wrote:
Q) A series of positive numbers P1, P2, P3, …. Pn, Pn+1 = Pn/3 for all values of n ≥ 1. what is the value of 2P4?
(1) P2 = 1/3
(2) P2 - P3 = 2/9

OA is D
need some help in going about this type of problem.


Okay, our first step is to define the formula for this series. As mentioned in my previous post, any value is obtained simply by taking the previous value and dividing that value by 3. That is what the P(n+1) = P(n)/3 means.

Okay, so if I want 2 * P(4), then I just need to know P(4) to be able to solve the question. Really, if I know any value in the sequence I could obtain P(4).

Statement 1 says that P(2) = 1/3. Sufficient. P(2) is 1/3, so P(3) is 1/9 (1/3 divided by 3). P(4) will be 1/9 divided by 3, or 1/27.

Statement 2 is only a little trickier:
Let's call P(2) = x
Because P(3) is P(2) divided by 3, P(3) = x/3.

Statement 2 says that P(2) - P(3) = 2/9. Thus x - x/3 = 2/9. 2/3x=2/9.
(3/2)2/3x = 2/9(3/2)
x=1/3
So we know that P(2) is 1/3. Sufficient.
Please note that we did not have to do much of this work for this statement. Once we had "x - x/3 = 2/9" we knew we could solve for x, which is standing in for P(2), and we should immediately have just said sufficient and moved on.

_________________
Jamie Nelson
ManhattanGMAT Instructor


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