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Arithmetic Series help (AS Level)

  1. Nov 18, 2006 #1
    I'm totally stuck on the following question and so I'd very very grateful if someone could please tell me how to work it out.

    The first three terms of an arithmetic series are (12-P), 2P and (4P-5) respectively, where P is a constant.

    (a) Find the value of P.

    (b) Show that the sixth term of the series is 50.

    (c) Find the sum of the first 15 terms of the series.

    (d) Find how many terms of the series have a value of less than 400.


    Thank you.

    Cathy
     
  2. jcsd
  3. Nov 18, 2006 #2

    quasar987

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    You have to show some work in order to get help.
     
  4. Nov 18, 2006 #3

    radou

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    Present us some of your work. Write down the expression for the general term of an arithmetic sequence, and everything should be more clear. Set up a few equations, and see where they'll bring you.

    Edit: late again. :smile:
     
  5. Nov 18, 2006 #4
    Before I posted I had written the following in response to part a:

    an = a1 + (n - 1)d

    d = 2p - (12 - p)

    d= (4p - 5) - 2p

    2p - (12 - p) = (4p - 5) - 2p

    4p = (4p - 5) + (12 - p)

    Then I got d = 7 but I got confused over the value of p.

    Am I working on the right lines?

    Thanks for replying by the way! :smile:

    Cathy
     
  6. Nov 18, 2006 #5

    radou

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    No, p should equal 7.
     
  7. Nov 18, 2006 #6
    Oh, okay, I get where I went wrong. Thank you.

    So, for part b do I just substitute in p to the equation to find d?

    Cathy
     
  8. Nov 18, 2006 #7

    radou

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    Yes, you do. I.e., d must equal a3 - a2, and a2 - a1, doesn't matter which difference you take.
     
  9. Nov 18, 2006 #8
    Thanks for your help.

    I've now worked out parts a, b and c but I'm still unsure how to do d. Any hints would be really appreciated.

    Cathy
     
  10. Nov 18, 2006 #9

    radou

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    Basically, you can solve part (d) unformally by trying to plug in different numbers n into an = a1 + (n - 1)*d and see which one is the greatest term which is less than 400.
     
  11. Nov 18, 2006 #10
    Okay, thanks very much! :biggrin:

    Cathy
     
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