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Math Help - Arithmetic Progression

  1. #1
    Senior Member slevvio's Avatar
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    Arithmetic Progression

    Hello I was having trouble with this problem and was wondering if anybody could help me with it. Show that if the sum of the first p terms of an arithmetic progression is equal to the sum of the first q terms, where p  \not= q, then the sum of the first p + q terms must be zero.

    I do not really know where to start; I came up with this but it is not helpful:

     \frac{p}{2}(2a + (p-1)d) = \frac{q}{2}(2a + (q-1)d)

    which of course shows that p = q even though it does not.

    Thanks and happy new year!
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  2. #2
    MHF Contributor

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    Look again.
    The sum of the first p terms is: \sum\limits_{k = 0}^{p - 1} {\left( {a + kd} \right)}  = pa + \frac{{\left( {p - 1} \right)p}d}{2}<br />
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  3. #3
    Lord of certain Rings
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    We will follow what you did,

    Quote Originally Posted by slevvio View Post
    I do not really know where to start; I came up with this but it is not helpful:
     \frac{p}{2}(2a + (p-1)d) = \frac{q}{2}(2a + (q-1)d)
    which of course shows that p = q even though it does not.
    Thanks and happy new year!
    Not really, it says p=q could be possible, but it doesn't say p=q is the only thing possible.
    What is the sum of first p+q terms??
    S = \frac{(p+q)(2a+(p+q-1)d)}2
    We want to show S=0 ,i.e 2a+(p+q-1)d = 0 --------------(*)
    Let us see why,
    \frac{p}{2}(2a + (p-1)d) = \frac{q}{2}(2a + (q-1)d)
    2ap +dp^2 - dp = 2aq + dq^2 - dq
    2a(p-q) +d(p^2 - q^2) -d(p-q) = 0
    (p-q)(2a + d(p+q)-d) = 0
    Since p \neq q we are forced to conclude
    2a+(p+q-1)d = 0 which is nothing but (*).

    All done
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