The answer is 315.

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We have AB = 14, BC = 16, and CD = 18.

Multiplying the first and third equations gives us $ABCD = 252$ , while multiplying all three equations gives us $A(B^{2})(C^{2})D = 252 \times 16$ . Therefore, we can say that $252^{2} = 252 \times 16AB$ . Solving the equation gets us $AB = 63/4$ , so $20AB$ is $\boxed{315}$ .