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# In a sequence of numbers in which each term after the first

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Senior Manager
Joined: 27 Aug 2007
Posts: 253

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In a sequence of numbers in which each term after the first [#permalink]

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25 Sep 2007, 06:32
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

In a sequence of numbers in which each term after the first term is 1 more than twice the preceding term, what is the fifth term?

(1) The first term is 1.
(2) The sixth term minus the fifth term is 32.

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VP
Joined: 08 Jun 2005
Posts: 1143

Kudos [?]: 254 [1], given: 0

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25 Sep 2007, 09:26
1
KUDOS
n
2n+1
2*(2n+1)+1 = 4n+3
2*(4n+3)+1 = 8n+7
2*(8n+7)+1 = 16n+15
2*(16n+15)+1 = 32n+31

statement 1

if n=1 then 16n+15 = 31

sufficient

statement 2

32n+31-(16n+15) = 32

32n+31-16n-15 = 32

16n+16 = 32

16n = 16

n = 1

if n=1 then 16n+15 = 31

sufficient

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Senior Manager
Joined: 27 Aug 2007
Posts: 253

Kudos [?]: 12 [0], given: 0

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25 Sep 2007, 09:36
Thank you KillerSquirrel !!!

I see where I did a mistake, Keep it UP!!!

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CEO
Joined: 29 Mar 2007
Posts: 2553

Kudos [?]: 528 [0], given: 0

Re: In a sequence of numbers... [#permalink]

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25 Sep 2007, 20:16
Ferihere wrote:
In a sequence of numbers in which each term after the first term is 1 more than twice the preceding term, what is the fifth term?

(1) The first term is 1.
(2) The sixth term minus the fifth term is 32.

Sequence:

n
1+2n
3+4n<------2(1+2n)+1
7+8n<------2(2(1+2n)+1)+1
15+16n<------2(2(2(1+2n)+1)+1)+1
31+32n<2> 16+16n=32 16n=16 ---> n=1

so 5th term:

15+16(1) = 31.

Ans D.

No reason to really do all that work though b/c a lot of room for error, just know u could have done it and save 3min =)

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Non-Human User
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Re: In a sequence of numbers in which each term after the first [#permalink]

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30 Aug 2017, 05:41
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Re: In a sequence of numbers in which each term after the first   [#permalink] 30 Aug 2017, 05:41
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# In a sequence of numbers in which each term after the first

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