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In a certain sequence a1, a2, ...an..., for n>1, each term is the sum

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Manager
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Joined: 30 May 2018
Posts: 61
GMAT 1: 620 Q42 V34
WE: Corporate Finance (Commercial Banking)
In a certain sequence a1, a2, ...an..., for n>1, each term is the sum  [#permalink]

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New post 06 Apr 2019, 22:22
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Question Stats:

53% (02:04) correct 47% (01:58) wrong based on 45 sessions

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In a certain sequence a1, a2, ...an..., for n>1, each term is the sum of all previous terms. If an=S, in terms of S, a(n+3)=?

(A) 3S
(B) 4S
(C) 6S
(D) 8S
(E) 9S
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Joined: 21 Feb 2019
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Location: Italy
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Re: In a certain sequence a1, a2, ...an..., for n>1, each term is the sum  [#permalink]

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New post 07 Apr 2019, 16:34
1
\(a_n = S\)
\(a_{n+1} = S + S = 2S\)
\(a_{n+2} = 2S + 2S = 4S\)
\(a_{n+3} = 4S + 4S = 8S\)
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Re: In a certain sequence a1, a2, ...an..., for n>1, each term is the sum  [#permalink]

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New post 22 Apr 2019, 01:02
lucajava wrote:
\(a_n = S\)
\(a_{n+1} = S + S = 2S\)
\(a_{n+2} = 2S + 2S = 4S\)
\(a_{n+3} = 4S + 4S = 8S\)


can you explain the logic behind this?
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Re: In a certain sequence a1, a2, ...an..., for n>1, each term is the sum  [#permalink]

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New post 22 Apr 2019, 01:45
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Ish1996 each term is the sum of all previous terms, so if \(a_n = S\), \(a_{n+1}\) is equal to all previous terms of \(a_n\), which give S as result, in addition to \(a_n\) itself, which is S.
Repeat this reasoning for the following ones and you will get the correct answer.

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Re: In a certain sequence a1, a2, ...an..., for n>1, each term is the sum   [#permalink] 22 Apr 2019, 01:45
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