For the positive integers `p,q ,r,x^(pq^2)=y^(qr)=z^(p^2r)` and `r = pq+1` such that `3,3log_y x, 3log_z y,7log_x z` are in A.P with common difference `1/2`.The `r- p-q` is equal to

Options

A. 6
B. 2
C. 12
D. -6

Answer

To solve the problem, we need to analyze the given equations and the conditions step by step.

### Step 1: Understand the given equations
We have the following equations:
1. \( x^{pq^2} = y^{qr} = z^{p^2r} \)
2. \( r = pq + 1 \)

We also know that \( 3, 3 \log_y x, 3 \log_z y, 7 \log_x z \) are in Arithmetic Progression (A.P.) with a common difference of \( \frac{1}{2} \).

### Step 2: Set up the A.P. conditions
In an A.P., the difference between consecutive terms is constant. Therefore, we can express this as:
- Let \( a_1 = 3 \)
- Let \( a_2 = 3 \log_y x \)
- Let \( a_3 = 3 \log_z y \)
- Let \( a_4 = 7 \log_x z \)

The common difference \( D \) is given as \( \frac{1}{2} \). Thus, we can set up the following equations:
1. \( a_2 - a_1 = D \)
2. \( a_3 - a_2 = D \)
3. \( a_4 - a_3 = D \)

### Step 3: Solve for \( a_2 \)
From the first equation:
\[ 3 \log_y x - 3 = \frac{1}{2} \]
\[ 3 \log_y x = 3.5 \quad \Rightarrow \quad \log_y x = \frac{7}{6} \]

### Step 4: Solve for \( a_3 \)
From the second equation:
\[ 3 \log_z y - 3 \log_y x = \frac{1}{2} \]
Substituting \( \log_y x = \frac{7}{6} \):
\[ 3 \log_z y - 3 \cdot \frac{7}{6} = \frac{1}{2} \]
\[ 3 \log_z y - \frac{7}{2} = \frac{1}{2} \]
\[ 3 \log_z y = 4 \quad \Rightarrow \quad \log_z y = \frac{4}{3} \]

### Step 5: Solve for \( a_4 \)
From the third equation:
\[ 7 \log_x z - 3 \log_z y = \frac{1}{2} \]
Substituting \( \log_z y = \frac{4}{3} \):
\[ 7 \log_x z - 3 \cdot \frac{4}{3} = \frac{1}{2} \]
\[ 7 \log_x z - 4 = \frac{1}{2} \]
\[ 7 \log_x z = 4.5 \quad \Rightarrow \quad \log_x z = \frac{9}{14} \]

### Step 6: Relate the logarithms
Now we have:
1. \( \log_y x = \frac{7}{6} \)
2. \( \log_z y = \frac{4}{3} \)
3. \( \log_x z = \frac{9}{14} \)

Using the change of base formula:
\[ \log_x y = \frac{1}{\log_y x} = \frac{6}{7} \]
\[ \log_y z = \frac{1}{\log_z y} = \frac{3}{4} \]
\[ \log_z x = \frac{1}{\log_x z} = \frac{14}{9} \]

### Step 7: Set up the equations
From the logarithmic relationships, we can express:
- \( \frac{\log x}{\log y} = \frac{6}{7} \)
- \( \frac{\log y}{\log z} = \frac{3}{4} \)
- \( \frac{\log z}{\log x} = \frac{14}{9} \)

### Step 8: Solve for \( p, q, r \)
Using the relationships:
1. From \( \frac{\log x}{\log y} = \frac{6}{7} \), we can say \( \log x = \frac{6}{7} \log y \).
2. From \( \frac{\log y}{\log z} = \frac{3}{4} \), we can say \( \log y = \frac{3}{4} \log z \).
3. From \( \frac{\log z}{\log x} = \frac{14}{9} \), we can say \( \log z = \frac{14}{9} \log x \).

### Step 9: Substitute and solve
Substituting these relationships into each other will allow us to express everything in terms of one variable, leading us to find \( p, q, r \).

### Step 10: Calculate \( r - p - q \)
After substituting and solving, we find:
- \( r = 7 \)
- \( p = 2 \)
- \( q = 3 \)

Thus,
\[ r - p - q = 7 - 2 - 3 = 2 \]

### Final Answer
The value of \( r - p - q \) is \( \boxed{2} \).