110k views
0 votes
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?

User Adrienne
by
5.2k points

1 Answer

3 votes

Answer: 503

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.

User Sinanspd
by
4.9k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.