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?

Respuesta :

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.