If x,y,z be three positive prime numbers. The progression in which `sqrt(x),sqrt(y),sqrt(z)` can be three terms (not necessarily consecutive) is

Options

A. AP
B. GP
C. HP
D. None of these

Explanation

To determine the type of progression in which \(\sqrt{x}, \sqrt{y}, \sqrt{z}\) can be three terms (where \(x, y, z\) are positive prime numbers), we will check if these terms can be in Arithmetic Progression (AP), Geometric Progression (GP), or Harmonic Progression (HP).

### Step 1: Check for Arithmetic Progression (AP)

In AP, the difference between consecutive terms is constant. We can express this as:
\[ \sqrt{y} - \sqrt{x} = d \quad \text{and} \quad \sqrt{z} - \sqrt{y} = d \]
This implies:
\[ \sqrt{y} - \sqrt{x} = \sqrt{z} - \sqrt{y} \]
Squaring both sides gives:
\[ y + x - 2\sqrt{xy} = z + y - 2\sqrt{yz} \]
Rearranging leads to:
\[ x - z = 2(\sqrt{xy} - \sqrt{yz}) \]
The left-hand side (LHS) is a difference of two primes, which is an integer, while the right-hand side (RHS) involves square roots. Since the LHS is rational and the RHS is irrational, they cannot be equal. Thus, \(\sqrt{x}, \sqrt{y}, \sqrt{z}\) cannot be in AP.

### Step 2: Check for Geometric Progression (GP)

In GP, the ratio of consecutive terms is constant. We can express this as:
\[ \frac{\sqrt{y}}{\sqrt{x}} = r \quad \text{and} \quad \frac{\sqrt{z}}{\sqrt{y}} = r \]
This implies:
\[ \sqrt{y}^2 = r^2 \cdot x \quad \text{and} \quad \sqrt{z}^2 = r^2 \cdot y \]
From this, we can express \(y\) and \(z\) in terms of \(x\):
\[ y = r^2 x \quad \text{and} \quad z = r^2 y = r^4 x \]
Since \(x, y, z\) are prime numbers, \(r^2\) must be such that both \(y\) and \(z\) remain prime. However, for any prime \(x\), \(r^2\) would need to be a rational number that results in \(y\) and \(z\) also being prime. This is not possible as it would imply that \(x, y, z\) share common factors, contradicting their primality. Thus, \(\sqrt{x}, \sqrt{y}, \sqrt{z}\) cannot be in GP.

### Step 3: Check for Harmonic Progression (HP)

In HP, the reciprocals of the terms are in AP. Thus, we check:
\[ \frac{1}{\sqrt{x}}, \frac{1}{\sqrt{y}}, \frac{1}{\sqrt{z}} \]
For these to be in AP:
\[ \frac{1}{\sqrt{y}} - \frac{1}{\sqrt{x}} = \frac{1}{\sqrt{z}} - \frac{1}{\sqrt{y}} \]
This leads to:
\[ \frac{\sqrt{y} - \sqrt{x}}{\sqrt{xy}} = \frac{\sqrt{z} - \sqrt{y}}{\sqrt{yz}} \]
Cross-multiplying gives:
\[ (\sqrt{y} - \sqrt{x}) \sqrt{yz} = (\sqrt{z} - \sqrt{y}) \sqrt{xy} \]
This equation again leads to a situation where we have irrational and rational components, similar to the previous cases. Therefore, \(\sqrt{x}, \sqrt{y}, \sqrt{z}\) cannot be in HP.

### Conclusion

Since we have shown that \(\sqrt{x}, \sqrt{y}, \sqrt{z}\) cannot be in AP, GP, or HP, the answer is:
**None of these.**