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Note that $a+b, a-b$ are consecutive terms of the arithmetical progression. Thus the difference of the progression is $-2b$ , and thus the third term $ab$ is equal to $a-3b$ , and the fourth term $\frac{a}{b}$ is equal to $a-5b$ .

Since $b \neq 0$ , we can multiply $b$ to both sides of the equation $\frac{a}{b} = a-5b$ , giving $a = ab - 5b^2$ . We also have $ab = a-3b$ . Substituting, we get:

$\begin{aligned} a &= (a-3b) - 5b^2 \\ 5b^2 + 3b &= 0 \\ b(5b+3) &= 0 \end{aligned}$

Since $b \neq 0$ , we have $b = \frac{-3}{5}$ . Substituting to $ab = a-3b$ , we obtain:

$\begin{aligned} a \cdot \frac{-3}{5} &= a - 3 \cdot \frac{-3}{5} \\ \frac{-8}{5} \cdot a &= \frac{9}{5} \\ a &= \frac{-9}{8} \end{aligned}$

Just to check, we can verify this: $a+b = \frac{-69}{40}, a-b = \frac{-21}{40}, ab = \frac{27}{40}, \frac{a}{b} = \frac{15}{8}$ indeed forms an arithmetical progression.

Thus $a+b = \frac{-69}{40}$ , giving $x = -69, y = 40$ , and $|x+y| = |-69+40| = \boxed{29}$ .