Math

posted by Matt

Penn writes a 2013-term arithmetic sequence of positive integers, and Teller writes a different 2013-term arithmetic sequence of integers. Teller's first term is the negative of Penn's first term. Each then finds the sum of terms in his sequence. If their sums are equal, then what is the smallest possible values of the first term in Penn's sequence?

  1. Steve

    If the sequences are P and T, and their differences are Pd and Td, then

    T1 = -P1
    2013/2 (2T1+2012Td) = 2013/2 (2P1 + 2012Pd)

    or,

    2013P1 + 2025078Pd = 2013T1 + 2025078Td
    4026P1 + 2025078(Pd-Td) = 0

    We know that since T1 < 0 and the sum of T is the same as the sum of P, that Td > Pd. What if Td = Pd+1? Then we have

    4026P1 = 2025078
    P1 = 503

    What if Td = Pd+k, with k>0?

    4026P1 = 2025078k
    P1 = 503k

    Let's check.
    P = 503,504,...,2515
    Sum = 2013/2 (503+2515) = 3037617

    T = -503,-501,...,3521
    Sum = 2013/2 (-503+3521) = 3037617

    cool problem.

  2. Billybob

    I got 503.....

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