Question

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?

Answer 1

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.

Answer 2

I got 503.....