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VP
Joined: 27 Dec 2004
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In a sequence, the first three values are the first three [#permalink]
 03 Jul 2005, 14:35
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"In a sequence, the first three values are the first three prime integers in ascending order, and each of the succeeding values is the sum of all the preceding values, what is the ratio of the Thirtieth value to the Twenty-fifth value in the sequence?"
A. 6/5
B. 32
C. 64
D. 128
E. 230
Please advise on a quicker way to solve there series problems. Than,s
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Director
Joined: 18 Apr 2005
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Re: PS- Quicker way to solve this [#permalink]
03 Jul 2005, 14:45
Folaa3 wrote: "In a sequence, the first three values are the first three prime integers in ascending order, and each of the succeeding values is the sum of all the preceding values, what is the ratio of the Thirtieth value to the Twenty-fifth value in the sequence?"
A. 6/5
B. 32
C. 64
D. 128
E. 230
Please advise on a quicker way to solve there series problems. Than,s
2 3 5 10 20 40 80 160 320 640 ...
N(30) = 10*2^(30 - 4)
N(25) = 10*2^(25 - 4)
so N(30)/N(25) = 2^5 = 32 or B
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VP
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Thanks sparky,
I used brute force and still arrived at 32 but it was in fact exhaustive. 
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Senior Manager
Joined: 17 Apr 2005
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Re: PS- Quicker way to solve this [#permalink]
04 Jul 2005, 07:36
Folaa3 wrote: "In a sequence, the first three values are the first three prime integers in ascending order, and each of the succeeding values is the sum of all the preceding values, what is the ratio of the Thirtieth value to the Twenty-fifth value in the sequence?"
A. 6/5
B. 32
C. 64
D. 128
E. 230
Please advise on a quicker way to solve there series problems. Than,s
B.
for n>3
nth term = sum(n-1) terms.
(n+1)th term = nth term + sum(n-1)terms
= nth term + nth term.
= 2 * nth terms.
Now we can go for the kill, GP with common ratio of 2.
HMTG.
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Current Student
Joined: 28 Dec 2004
Posts: 3437
Location: New York City
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Re: PS- Quicker way to solve this [#permalink]
04 Jul 2005, 10:35
good one sparky...I took me 30 min to come up with an equation 
too much partying lastnight...
I just looked at the stem again and now it took me 3 mins to realize the sequence...
sparky wrote: Folaa3 wrote: "In a sequence, the first three values are the first three prime integers in ascending order, and each of the succeeding values is the sum of all the preceding values, what is the ratio of the Thirtieth value to the Twenty-fifth value in the sequence?"
A. 6/5
B. 32
C. 64
D. 128
E. 230
Please advise on a quicker way to solve there series problems. Than,s 2 3 5 10 20 40 80 160 320 640 ... N(30) = 10*2^(30 - 4) N(25) = 10*2^(25 - 4) so N(30)/N(25) = 2^5 = 32 or B |
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Manager
Joined: 10 Sep 2005
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am still clueless...how did you derive at the equation
N(30) = 10*2^(30 - 4)
N(25) = 10*2^(25 - 4)
[/quote]
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Manager
Joined: 06 Oct 2005
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Sushom, look at the sequence for the first few numbers
2, 3, 5, 10, 20, 40, 80
Starting with the fifth member of the sequence it is 10*2^1, the sixth is 10*2^2, the seventh is 10*2^3......see the pattern in relation to the exponent and sequence position?
starting with the fifth member x=10*2^n-4
therefore when n=30 x=10*2^26 and when n=25 x=10*2^21
B.
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