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20 September, 02:13

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 the terms in his sequence. If their sums are equal, then what is the smallest possible value of the first term in Penn's sequence?

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  1. 20 September, 02:17
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    Answer: 503

    Step-by-step explanation:

    Penn: n = 2013

    a₁ = x

    r = rp

    Teller n = 2013

    a₁ = - x

    r = rt

    Sum of terms arithmetic sequence: S = (a₁ + aₙ) n/2

    Spenn = (x + a₂₀₁₃).2013/2

    Steller = (-x + a₂₀₁₃).2013/2

    Spenn = Steller

    (x + ap₂₀₁₃).2013/2 = (-x + at₂₀₁₃).2013/2

    2x = at₂₀₁₃ - ap₂₀₁₃

    General term of arithmetic sequence: aₙ = a₁ + (n-1) r

    ap₂₀₁₃ = x + (2013-1) rp = x + 2012rp

    at₂₀₁₃ = - x + (2013-1) rt = - x + 2012rt

    2x = at₂₀₁₃ - ap₂₀₁₃

    2x = - x + 2012rt - (x + 2012rp)

    2x = - x + 2012rt - x - 2012rp

    2x = - 2x + 2012rt - 2012rp

    4x = 2012rt - 2012rp

    4x = 2012 (rt-rp)

    x = 2012 (rt-rp) / 4

    x = 503 (rt-rp)

    As rt - rp cannot be 0, and we know that x is positive, the smallest number for (rt-rp) is 1, so the smallest number for x is 503.
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