Let `a, x, b` in A.P, `a, y, b` in GP `a, z, b` in HP where `a` and `b` are disnict positive real numbers. If `x = y + 2` and `a = 5z`, then which of the following is/are `COR RECT` ?

Options

A. `y^(2) = zx`
B. `x gt y gt z`
C. `a = 9 , b = 1`
D. `a = (1)/(4), b = -(9)/(4)`

Explanation

To solve the problem step by step, we will use the properties of Arithmetic Progression (A.P.), Geometric Progression (G.P.), and Harmonic Progression (H.P.) as given in the question.

### Step 1: Understanding the Progressions
1. **A.P. Condition**: Since \(a, x, b\) are in A.P., we can write:
\[ 2x = a + b \quad \text{(1)} \]

2. **G.P. Condition**: Since \(a, y, b\) are in G.P., we can write:
\[ y^2 = ab \quad \text{(2)} \]

3. **H.P. Condition**: Since \(a, z, b\) are in H.P., we can write:
\[ z = \frac{2ab}{a + b} \quad \text{(3)} \]

### Step 2: Using Given Relationships
We are given:
- \(x = y + 2\) (4)
- \(a = 5z\) (5)

### Step 3: Substitute for \(z\) in terms of \(a\) and \(b\)
From equation (3):
\[ z = \frac{2ab}{a + b} \]
Substituting this into equation (5):
\[ a = 5\left(\frac{2ab}{a + b}\right) \]
Multiplying both sides by \(a + b\):
\[ a(a + b) = 10ab \]
Expanding and rearranging gives:
\[ a^2 + ab - 10ab = 0 \implies a^2 - 9ab = 0 \]
Factoring out \(a\):
\[ a(a - 9b) = 0 \]
Since \(a\) and \(b\) are distinct positive real numbers, we have:
\[ a = 9b \quad \text{(6)} \]

### Step 4: Substitute \(a\) into the A.P. Condition
Substituting equation (6) into equation (1):
\[ 2x = 9b + b = 10b \implies x = 5b \quad \text{(7)} \]

### Step 5: Substitute \(a\) into the G.P. Condition
Substituting equation (6) into equation (2):
\[ y^2 = (9b)b = 9b^2 \implies y = 3b \quad \text{(8)} \]

### Step 6: Verify the Relationship Between \(x\) and \(y\)
Using equation (4):
\[ x = y + 2 \implies 5b = 3b + 2 \implies 2b = 2 \implies b = 1 \]
Now substituting \(b = 1\) back into equations (6), (7), and (8):
- From (6): \(a = 9b = 9 \times 1 = 9\)
- From (7): \(x = 5b = 5 \times 1 = 5\)
- From (8): \(y = 3b = 3 \times 1 = 3\)

### Step 7: Find \(z\)
Substituting \(a\) and \(b\) into equation (3):
\[ z = \frac{2ab}{a + b} = \frac{2 \times 9 \times 1}{9 + 1} = \frac{18}{10} = \frac{9}{5} \]

### Final Values
- \(a = 9\)
- \(b = 1\)
- \(x = 5\)
- \(y = 3\)
- \(z = \frac{9}{5}\)

### Conclusion
The values satisfy the conditions of A.P., G.P., and H.P. Therefore, the correct statements regarding the relationships among \(x\), \(y\), and \(z\) can be verified.