Key Concepts and Formulas

• Arithmetic Progression (A.P.): A sequence of numbers such that the difference between the consecutive terms is constant. For three terms in A.P., they can be represented as $a-d, a, a+d$.
• Sum of an A.P. (3 terms): $(a-d) + a + (a+d) = 3a$.
• Product of an A.P. (3 terms): $(a-d) \times a \times (a+d) = a(a^2 - d^2)$.
• Algebraic Manipulation: Solving systems of equations and simplifying expressions.

Step-by-Step Solution

Step 1: Represent the sets A and B using A.P. properties. Let set A contain three numbers in A.P. with common difference $d$. We can represent these numbers as $a_1 - d, a_1, a_1 + d$. Let set B contain three numbers in A.P. with common difference $D$. We can represent these numbers as $a_2 - D, a_2, a_2 + D$. We are given that $D = d + 3$ and $d > 0$.

Step 2: Use the given sum of elements to find the middle terms of the A.P.s. The sum of the elements in A is 36: $(a_1 - d) + a_1 + (a_1 + d) = 36$ $3a_1 = 36$ $a_1 = 12$ The sum of the elements in B is 36: $(a_2 - D) + a_2 + (a_2 + D) = 36$ $3a_2 = 36$ $a_2 = 12$ This shows that the middle term of both A.P.s is 12.

Step 3: Express the products p and q in terms of the common differences. The product of the elements in A is $p$: $p = (a_1 - d)(a_1)(a_1 + d)$ Substituting $a_1 = 12$: $p = (12 - d)(12)(12 + d) = 12(12^2 - d^2) = 12(144 - d^2)$ The product of the elements in B is $q$: $q = (a_2 - D)(a_2)(a_2 + D)$ Substituting $a_2 = 12$: $q = (12 - D)(12)(12 + D) = 12(12^2 - D^2) = 12(144 - D^2)$

Step 4: Use the given ratio $\frac{p+q}{p-q}=\frac{19}{5}$ to establish a relationship between p and q. Cross-multiplying the given ratio: $5(p + q) = 19(p - q)$ $5p + 5q = 19p - 19q$ $5q + 19q = 19p - 5p$ $24q = 14p$ Dividing both sides by 2: $12q = 7p$

Step 5: Substitute the expressions for p and q into the relationship $12q = 7p$. $12 [12(144 - D^2)] = 7 [12(144 - d^2)]$ Divide both sides by 12: $12(144 - D^2) = 7(144 - d^2)$

Step 6: Substitute $D = d+3$ into the equation and simplify. $12[144 - (d+3)^2] = 7(144 - d^2)$ Expand $(d+3)^2$: $12[144 - (d^2 + 6d + 9)] = 7(144 - d^2)$ $12[144 - d^2 - 6d - 9] = 7(144 - d^2)$ $12[135 - d^2 - 6d] = 7(144 - d^2)$ Distribute the constants: $1620 - 12d^2 - 72d = 1008 - 7d^2$ Rearrange the terms to form a quadratic equation: $0 = (12d^2 - 7d^2) + 72d + (1008 - 1620)$ $0 = 5d^2 + 72d - 612$

Step 7: Solve the quadratic equation for d. We can solve $5d^2 + 72d - 612 = 0$ by factoring or using the quadratic formula. Let's try factoring: We look for two numbers that multiply to $5 \times (-612) = -3060$ and add up to 72. These numbers are 102 and -30. $5d^2 + 102d - 30d - 612 = 0$ $d(5d + 102) - 6(5d + 102) = 0$ $(d - 6)(5d + 102) = 0$ The possible values for $d$ are $d = 6$ or $d = -\frac{102}{5}$. Since we are given that $d > 0$, we choose $d = 6$.

Step 8: Calculate D and then p and q. Now that we have $d=6$, we can find $D$: $D = d + 3 = 6 + 3 = 9$ Now calculate $p$: $p = 12(144 - d^2) = 12(144 - 6^2) = 12(144 - 36) = 12(108)$ $p = 1296$ Now calculate $q$: $q = 12(144 - D^2) = 12(144 - 9^2) = 12(144 - 81) = 12(63)$ $q = 756$

Step 9: Calculate p - q. $p - q = 1296 - 756$ $p - q = 540$

Common Mistakes & Tips

• When expanding $(d+3)^2$, ensure that all terms are correctly accounted for: $d^2 + 6d + 9$.
• Carefully check the arithmetic when solving the quadratic equation, especially the signs and calculations involved in factoring or the quadratic formula.
• Always remember to use the given condition $d > 0$ to select the correct value of $d$.

Summary

The problem involves understanding the properties of arithmetic progressions, specifically how to represent terms and calculate their sum and product. By setting up equations based on the given information about the sums, products, and the relationship between the common differences, we derived a quadratic equation for $d$. Solving this equation and using the condition $d>0$ allowed us to find the specific values of $d$ and $D$, which in turn enabled us to calculate the products $p$ and $q$. Finally, the difference $p-q$ was computed.

The final answer is $\boxed{540}$.