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

Let the elements of set \( A \) be \( a - d, a, a + d \) (since they are in A.P.) and the elements of set \( B \) be \( b - D, b, b + D \). 
The sum of the elements of set \( A \) is given by: \[ (a - d) + a + (a + d) = 3a = 36 \quad \Rightarrow \quad a = 12 \] The product of the elements of set \( A \) is: \[ (a - d) \cdot a \cdot (a + d) = a(a^2 - d^2) = p \] \[ 12 \cdot (12^2 - d^2) = p \quad \Rightarrow \quad 12(144 - d^2) = p \] Similarly, the sum of the elements of set \( B \) is: \[ (b - D) + b + (b + D) = 3b = 36 \quad \Rightarrow \quad b = 12 \] The product of the elements of set \( B \) is: \[ (b - D) \cdot b \cdot (b + D) = b(b^2 - D^2) = q \] \[ 12 \cdot (12^2 - D^2) = q \quad \Rightarrow \quad 12(144 - D^2) = q \] We are given that \( D = d + 3 \), so substitute \( D = d + 3 \) into the equation for \( q \): \[ q = 12(144 - (d + 3)^2) \] Now, we are given the relation: \[ \frac{p + q}{p - q} = \frac{19}{5} \] Substitute the expressions for \( p \) and \( q \) into this relation, and solve for \( p - q \). After solving, we get \( p - q = 540 \). 
Thus, the correct answer is \( 540 \).