Question

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 A.P's in $ A $ and $ B $, respectively, such that $ D = d + 3 $, $ d>0 $. If $ \frac{p+q}{p-q} = \frac{19}{5} $, then $ p - q $ is equal to:

• A. 600
• B. 450
• C. 630
• D. 540

Hint:

For problems involving sums and products of terms in an A.P., use the standard formulas for sum and product in A.P. and substitute the given values accordingly.

Answer 1

The Correct Option is D

Given set \( A \) with elements \( a - d, a, a + d \) (an arithmetic progression) and set \( B \) with elements \( b - D, b, b + D \).
The sum of elements in \( A \) is \( (a - d) + a + (a + d) = 3a \). Since \( 3a = 36 \), \( a = 12 \).
The product of elements in \( A \) is \( (a - d) \cdot a \cdot (a + d) = a(a^2 - d^2) = p \). Substituting \( a = 12 \), we get \( 12(12^2 - d^2) = p \), which simplifies to \( 12(144 - d^2) = p \).
Similarly, the sum of elements in \( B \) is \( (b - D) + b + (b + D) = 3b \). Since \( 3b = 36 \), \( b = 12 \).
The product of elements in \( B \) is \( (b - D) \cdot b \cdot (b + D) = b(b^2 - D^2) = q \). Substituting \( b = 12 \), we get \( 12(12^2 - D^2) = q \), which simplifies to \( 12(144 - D^2) = q \).
Given \( D = d + 3 \), substitute this into the equation for \( q \): \( q = 12(144 - (d + 3)^2) \).
The relation \( \frac{p + q}{p - q} = \frac{19}{5} \) is provided. Substituting the expressions for \( p \) and \( q \) and solving for \( p - q \) yields \( p - q = 540 \). 
Therefore, the result is \( 540 \).