Consider two sets A and B, each containing three numbers in A.P. Let the sum and the product of the elements of A be 36 and p respectively and the sum and the product of the elements of B be 36 and q respectively. Let d and D be the common differences of AP′ s in A and B respectively such that D=d+3, d>0. If (p+q)/(p-q)=19/5, then p-q is equal to

Solution

Maths sir · 4 steps

Write each A.P. as a-d, a, a+d: the sum fixes the middle term, and the product is easy.

1. Middle term first. Take A as 12-d, 12, 12+d, because their sum is 3 × 12 = 36. In the same way, B is 12-D, 12, 12+D. So p = 12(144-d²) and q = 12(144-D²).
2. Use the ratio. Cross-multiply (p+q)/(p-q) = 19/5 to get 5p + 5q = 19p - 19q. That gives 24q = 14p, so 7p = 12q. The common factor 12 cancels, which leaves 7(144-d²) = 12(144-D²).

   12(d+3)² - 7d² = 720 ⇒ 5d² + 72d - 612 = 0

3. Solve for d. The discriminant is 72² + 4 × 5 × 612 = 17424 = 132², so d = (-72 + 132)/10 = 6. The other root is negative, and d > 0 rules it out. Then D = 9.
4. So. p = 12(144 - 36) = 1296 and q = 12(144 - 81) = 756. So p - q = 540. Check: 2052/540 = 19/5.

The idea to keep. For three terms in A.P., write them as a-d, a, a+d. The sum gives a at once, and the product becomes a(a²-d²).

Answer from the NTA final answer key. Solution by PrepWiser.