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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.

    BiP
     
  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
    Thanks!

    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!

    Genius!

    BiP
     
  5. Jul 1, 2012 #4

    Mark44

    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.

    BiP
     
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