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Interesting problem involving arithmetic progression

  1. Jul 1, 2012 #1
    I just came up with a problem I hope you will find interesting, but I can't seem it solve it myself. I thought of induction as some guide, but am not sure how to proceed.

    There are N terms in some finite arithmetic progression. Two of those terms are equal to 3. Prove that all terms in this progression also equal 3.

  2. jcsd
  3. Jul 1, 2012 #2

    Simon Bridge

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    You are making life too hard for yourself :)

    An arithmetic progression is a sequence of numbers such that the difference of any two successive members of the sequence is a constant.

    If term xi=a and xi+n=b then the difference between successive terms Δx must be Δx=(b-a)/n.

    If a=b then Δx must be ...
    Last edited: Jul 1, 2012
  4. Jul 1, 2012 #3

    So 2=2+n*d
    0 = n*d
    d = 0, since n is not 0.

    Since d=0, every term must equal the ith term, which happens to be 2!


  5. Jul 1, 2012 #4


    Staff: Mentor

    I'm assuming that the work above is related to the problem in the original post. The i-th term can't be 2 if you're proving that all the terms in the progression are 3.
  6. Jul 1, 2012 #5
    Sorry my mistake, I meant to write 3.

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