What is the fourth term of an arithmetic sequence with specific given terms?

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Discussion Overview

The discussion revolves around determining the fourth term of an arithmetic sequence given the first three terms expressed in terms of a variable \( p \). Participants explore whether the sequence is indeed arithmetic and how to compute the fourth term based on the value of \( p \).

Discussion Character

  • Homework-related
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant presents the problem of finding the fourth term of an arithmetic sequence defined by the terms \( p, 2p+6, \) and \( 5p-12 \).
  • Another participant argues that the sequence is not an arithmetic progression (AP) unless \( p = 12 \), providing a calculation to support this claim.
  • A later reply confirms that if \( p = 12 \), the difference between consecutive terms is 18, and uses the formula for the \( n \)-th term of an AP to calculate the fourth term as 66.
  • There is a challenge to the assertion that the sequence is not an AP, with one participant insisting that it is, given the specific value of \( p \).

Areas of Agreement / Disagreement

Participants disagree on whether the sequence is an arithmetic progression for all values of \( p \) or only when \( p = 12 \). The discussion remains unresolved regarding the generality of the sequence's arithmetic nature.

Contextual Notes

The discussion highlights the dependence on the value of \( p \) and the conditions under which the sequence can be classified as arithmetic. There are unresolved mathematical steps regarding the implications of different values of \( p \).

sabanation12
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Our 8th grade math counts team met today and I didnt know how to do this problem:

The first three terms of an arithmetic sequence are p, 2p+6, and 5p-12. What is the 4th term of this sequence?

Please explain how to do this.

Arigato!
 
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please help me
 
Its not an arithmetic progression... and don't bump. In an AP the difference between consecutive terms is constant (i.e. t1-t0=t2-t1 and so forth.) In this case (2p+6)-p=5p-12-(2p+6)
p+6=3p-18
2p=24
p=12
Only true when p=12. For all other cases it is not an AP.
 
so, is that the solution? p=12 and a difference of 18 between two consecutive numbers?

2p-12 <- simplifies to p, of course, if you know p=12
2p+6
5p-12
5p+6
8p-12
8p+6
.
.
.
 
hyurnat4 said:
Its not an arithmetic progression... and don't bump. In an AP the difference between consecutive terms is constant (i.e. t1-t0=t2-t1 and so forth.) In this case (2p+6)-p=5p-12-(2p+6)
p+6=3p-18
2p=24
p=12
Only true when p=12. For all other cases it is not an AP.

No I believe you are wrong, it IS an arithmetic sequence. Let me explain...

P does equal 12, and the difference between them is 18, so:

an = a1 + (n-1) * d

Plugging in numbers:

an = 12 + (4-1) * 18

so the fourth number is 66

Thanks for your help anyways :)
 

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