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(D) Let the elements of set $A$ be $(a-d, a, a+d)$. The sum is $3a = 36$,so $a = 12$. The product is $p = a(a^2 - d^2) = 12(144 - d^2)$.Let the elemen

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$540$

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(D) Let the elements of set $A$ be $(a-d, a, a+d)$. The sum is $3a = 36$,so $a = 12$. The product is $p = a(a^2 - d^2) = 12(144 - d^2)$.Let the elements of set $B$ be $(b-D, b, b+D)$. The sum is $3b = 36$,so $b = 12$. The product is $q = b(b^2 - D^2) = 12(144 - D^2)$.Given $\frac{p+q}{p-q} = \frac{19}{5}$. By componendo and dividendo,$\frac{p}{q} = \frac{19+5}{19-5} = \frac{24}{14} = \frac{12}{7}$.Substituting $p$ and $q$: $\frac{12(144-d^2)}{12(144-D^2)} = \frac{12}{7}$.Since $D = d+3$,$D^2 = (d+3)^2 = d^2 + 6d + 9$.$\frac{144-d^2}{144-(d^2+6d+9)} = \frac{12}{7} \implies \frac{144-d^2}{135-d^2-6d} = \frac{12}{7}$.$7(144-d^2) = 12(135-d^2-6d) \implies 1008 - 7d^2 = 1620 - 12d^2 - 72d$.$5d^2 + 72d - 612 = 0$. Solving for $d$ using the quadratic formula: $d = \frac{-72 \pm \sqrt{5184 - 4(5)(-612)}}{10} = \frac{-72 \pm \sqrt{5184 + 12240}}{10} = \frac{-72 \pm \sqrt{17424}}{10} = \frac{-72 \pm 132}{10}$.Since $d > 0$,$d = \frac{60}{10} = 6$. Then $D = 6+3 = 9$.$p - q = 12(144 - d^2) - 12(144 - D^2) = 12(D^2 - d^2) = 12(81 - 36) = 12(45) = 540$.

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