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# An arithmetic sequence of numbers begins p, 2p + 2, 4p + 3... What is

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Joined: 26 Nov 2017
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An arithmetic sequence of numbers begins p, 2p + 2, 4p + 3... What is  [#permalink]

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Updated on: 10 Feb 2018, 10:35
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85% (hard)

Question Stats:

43% (02:07) correct 57% (02:14) wrong based on 77 sessions

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An arithmetic sequence of numbers begins p, 2p + 2, 4p + 3... What is the 13th member of this sequence?

A. 37
B. 40
C. 26p + 13
D. 53
E. 129p + 13

Originally posted by henrymba2021 on 10 Feb 2018, 09:11.
Last edited by Bunuel on 10 Feb 2018, 10:35, edited 1 time in total.
Renamed the topic and edited the question.
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Re: An arithmetic sequence of numbers begins p, 2p + 2, 4p + 3... What is  [#permalink]

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10 Feb 2018, 11:42
1
henrymba2021 wrote:
An arithmetic sequence of numbers begins p, 2p + 2, 4p + 3... What is the 13th member of this sequence?

A. 37
B. 40
C. 26p + 13
D. 53
E. 129p + 13

We are given p,2p + 2, and 4p + 3.... as the sequence.
If p = 1, the sequence becomes 1,4,7.....

Here First term(a) = 1, Common difference(d) = 3

Therefore, the 13th member of this sequence must be $$a + 12d = 1 + 12*3 = 1 + 36 = 37$$ (Option A)
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Re: An arithmetic sequence of numbers begins p, 2p + 2, 4p + 3... What is  [#permalink]

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10 Feb 2018, 22:25
1
pushpitkc wrote:
henrymba2021 wrote:
An arithmetic sequence of numbers begins p, 2p + 2, 4p + 3... What is the 13th member of this sequence?

A. 37
B. 40
C. 26p + 13
D. 53
E. 129p + 13

We are given p,2p + 2, and 4p + 3.... as the sequence.
If p = 1, the sequence becomes 1,4,7.....

Here First term(a) = 1, Common difference(d) = 3

Therefore, the 13th member of this sequence must be $$a + 12d = 1 + 12*3 = 1 + 36 = 37$$ (Option A)

hi pushpitkc
why you have chosen P = 1 ...???
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An arithmetic sequence of numbers begins p, 2p + 2, 4p + 3... What is  [#permalink]

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10 Feb 2018, 22:48
1
sobby wrote:
pushpitkc wrote:
henrymba2021 wrote:
An arithmetic sequence of numbers begins p, 2p + 2, 4p + 3... What is the 13th member of this sequence?

A. 37
B. 40
C. 26p + 13
D. 53
E. 129p + 13

We are given p,2p + 2, and 4p + 3.... as the sequence.
If p = 1, the sequence becomes 1,4,7.....

Here First term(a) = 1, Common difference(d) = 3

Therefore, the 13th member of this sequence must be $$a + 12d = 1 + 12*3 = 1 + 36 = 37$$ (Option A)

hi pushpitkc
why you have chosen P = 1 ...???

Hey sobby

When I tried p=2 or 3, the emerging sequence was not having a common difference.
An arithmetic sequence has a common difference. That is the reason I chose p=1
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Joined: 02 Aug 2009
Posts: 7967
An arithmetic sequence of numbers begins p, 2p + 2, 4p + 3... What is  [#permalink]

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10 Feb 2018, 23:44
3
1
henrymba2021 wrote:
An arithmetic sequence of numbers begins p, 2p + 2, 4p + 3... What is the 13th member of this sequence?

A. 37
B. 40
C. 26p + 13
D. 53
E. 129p + 13

Hi..
An arithmetic sequence has same difference between consecutive terms..
So 2p+2-p=4p+3-(2p+2)....
.p+2=2p+1,
So 2p-p=2-1......p=1

Sequence is p, 2p+2, 4p+3.........
Difference in each term = 2p+2-p=p+2=1+2=3..
13 the term = 1+12*3=37
A
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Re: An arithmetic sequence of numbers begins p, 2p + 2, 4p + 3... What is  [#permalink]

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16 Sep 2019, 17:04
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Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: An arithmetic sequence of numbers begins p, 2p + 2, 4p + 3... What is   [#permalink] 16 Sep 2019, 17:04
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