There are two sets A and B each of which consists of three numbers in GP whose product is 64 and R and r are the common ratios sich that `R=r+2`. If `(p)/(q)=(3)/(2)`, where p and q are sum of numbers taken two at a time respectively in the two sets.
The value of q is

Options

A. 54
B. 56
C. 58
D. 60

Explanation

To solve the problem, we need to follow these steps:

### Step 1: Understand the Given Information
We have two sets A and B, each consisting of three numbers in a geometric progression (GP). The product of the numbers in each set is 64. The common ratios for sets A and B are denoted as R and r, respectively, with the relationship \( R = r + 2 \).

### Step 2: Express the Numbers in GP
Let the three numbers in set A be:
- \( a/r \)
- \( a \)
- \( ar \)

The product of these numbers is given by:
\[ \frac{a}{r} \cdot a \cdot ar = a^3 = 64 \]
Thus, we find:
\[ a = 4 \]

### Step 3: Find the Common Ratios
Now, substituting \( a = 4 \) into the expressions for the numbers in set A, we have:
- \( \frac{4}{R} \)
- \( 4 \)
- \( 4R \)

For set B, the numbers will be:
- \( \frac{4}{r} \)
- \( 4 \)
- \( 4r \)

### Step 4: Calculate P and Q
Now, we calculate \( P \) and \( Q \), where \( P \) is the sum of products taken two at a time from set A:
\[ P = \left(\frac{4}{R} \cdot 4\right) + \left(4 \cdot 4R\right) + \left(\frac{4}{R} \cdot 4R\right) = \frac{16}{R} + 16R + 16 \]
Similarly, for set B:
\[ Q = \left(\frac{4}{r} \cdot 4\right) + \left(4 \cdot 4r\right) + \left(\frac{4}{r} \cdot 4r\right) = \frac{16}{r} + 16r + 16 \]

### Step 5: Set Up the Ratio
According to the problem, we have:
\[ \frac{P}{Q} = \frac{3}{2} \]
Substituting the expressions for \( P \) and \( Q \):
\[ \frac{\frac{16}{R} + 16R + 16}{\frac{16}{r} + 16r + 16} = \frac{3}{2} \]

### Step 6: Simplify the Equation
We can simplify the equation by multiplying both sides by \( 2 \left(\frac{16}{r} + 16r + 16\right) \):
\[ 2\left(\frac{16}{R} + 16R + 16\right) = 3\left(\frac{16}{r} + 16r + 16\right) \]

### Step 7: Substitute R in Terms of r
Since \( R = r + 2 \), we can substitute \( R \) in the equation:
\[ 2\left(\frac{16}{r + 2} + 16(r + 2) + 16\right) = 3\left(\frac{16}{r} + 16r + 16\right) \]

### Step 8: Solve the Equation
After substituting and simplifying, we will arrive at a cubic equation in terms of \( r \):
\[ r^3 + 5r^2 - 7r = 0 \]
Factoring gives us:
\[ (r - 2)(r^2 + 7r + 3) = 0 \]
Thus, \( r = 2 \) is a solution.

### Step 9: Find the Value of Q
Substituting \( r = 2 \) back into the expression for \( Q \):
\[ Q = \frac{16}{2} + 16(2) + 16 = 8 + 32 + 16 = 56 \]

Thus, the value of \( q \) is:
\[ \boxed{56} \]