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

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henrymba2021
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An arithmetic sequence of numbers begins p, 2p + 2, 4p + 3... What is [#permalink] New post   Updated on: 10 Feb 2018, 10:35
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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
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A

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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chetan2u
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An arithmetic sequence of numbers begins p, 2p + 2, 4p + 3... What is [#permalink] New post   10 Feb 2018, 23:44
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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] New post   10 Feb 2018, 11:42
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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] New post   10 Feb 2018, 22:25
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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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pushpitkc
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An arithmetic sequence of numbers begins p, 2p + 2, 4p + 3... What is [#permalink] New post   10 Feb 2018, 22:48
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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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Re: An arithmetic sequence of numbers begins p, 2p + 2, 4p + 3... What is [#permalink] New post   07 Sep 2023, 22:29
Given: An arithmetic sequence of numbers begins p, 2p + 2, 4p + 3.

Asked: What is the 13th member of this sequence?

Common difference = d = (2p + 2) - p = (4p+3) - (2p+2)
p + 2 = 2p + 1
p = 1

Arithmetic sequence = 1, 4, 7...
a = 1
d = 3
n = 13

\(t_13 = a + (n-1)d = 1 + 12*3 = 37\)

IMO A
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Re: An arithmetic sequence of numbers begins p, 2p + 2, 4p + 3... What is [#permalink]
07 Sep 2023, 22:29
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