Can you find pythagorean triples in which the smallest number is 9, 11, and 13?

18  

Step-by-step explanation:

The first four terms of your sequence are

a₁ = 34

a₂ = 30

a₃ = 26

a₄ = 22

If the sum of all the terms is zero, there must be a set of negative terms corresponding to the positive ones.

That is, the last four terms must be -22, -26, -30, -34.

So, one strategy is to find the number of positive terms and multiply by 2.

Each term (aₙ) in your sequence is 4 units less than the preceding term (aₙ₋₁) .

The explicit formula is

aₙ= 34 - 4(n - 1)

Assume aₙ = 0. Then

We can't have a fractional number of terms, so  

n = 9 (that is, the last positive number is 2).

There must be a corresponding sequence of nine negative numbers.

Your complete sequence has 18 terms.