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Arithmetic Sequence

  1. Nov 21, 2011 #1
    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!
     
  2. jcsd
  3. Nov 21, 2011 #2
    please help me
     
  4. Nov 22, 2011 #3
    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 [tex](2p+6)-p=5p-12-(2p+6)[/tex]
    [tex]p+6=3p-18[/tex]
    [tex]2p=24[/tex]
    [tex]p=12[/tex]
    Only true when p=12. For all other cases it is not an AP.
     
  5. Nov 22, 2011 #4
    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
    .
    .
    .
     
  6. Nov 22, 2011 #5
    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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