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Let $a,b$ be positive integers. Prove that the set $$C = \left\{(ax-by): 0 < x < b, x+y = \left\lfloor\dfrac{a+b}{2}\right\rfloor\right\}$$ where $x$ and $y$ are positive integers is an arithmetic progression of length $b-1$ with difference $a+b$.

I didn't see how to use the condition that $x+y = \left\lfloor\dfrac{a+b}{2}\right\rfloor$ in order to solve the question. How can we use it to show it is an arithmetic progression?

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Hint:

$$ax-by=(a+b)x-b(x+y)$$

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