If the number x,y,z are in H.P. , then `sqrt(yz)/(sqrt(y)+sqrt(z)),sqrt(xz)/(sqrt(x)+sqrt(z)),sqrt(xy)/(sqrt(x)+sqrt(y))` are in

Options

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

Explanation

To solve the problem, we need to prove that the quantities

\[ A = \frac{\sqrt{yz}}{\sqrt{y} + \sqrt{z}}, \quad B = \frac{\sqrt{xz}}{\sqrt{x} + \sqrt{z}}, \quad C = \frac{\sqrt{xy}}{\sqrt{x} + \sqrt{y}} \]

are in arithmetic progression (A.P.) given that \(x, y, z\) are in harmonic progression (H.P.).

### Step 1: Understanding Harmonic Progression
If \(x, y, z\) are in H.P., then their reciprocals \( \frac{1}{x}, \frac{1}{y}, \frac{1}{z} \) are in A.P. This means:

\[ \frac{1}{y} - \frac{1}{x} = \frac{1}{z} - \frac{1}{y} \]

### Step 2: Expressing A, B, and C
We can express \(A\), \(B\), and \(C\) in terms of their reciprocals:

\[ A = \frac{\sqrt{yz}}{\sqrt{y} + \sqrt{z}} = \frac{1}{\frac{1}{\sqrt{y}} + \frac{1}{\sqrt{z}}} \]

\[ B = \frac{\sqrt{xz}}{\sqrt{x} + \sqrt{z}} = \frac{1}{\frac{1}{\sqrt{x}} + \frac{1}{\sqrt{z}}} \]

\[ C = \frac{\sqrt{xy}}{\sqrt{x} + \sqrt{y}} = \frac{1}{\frac{1}{\sqrt{x}} + \frac{1}{\sqrt{y}}} \]

### Step 3: Finding the Relationship
To show that \(A\), \(B\), and \(C\) are in A.P., we need to show that:

\[ 2B = A + C \]

This can be rewritten as:

\[ B - A = C - B \]

### Step 4: Calculate \(B - A\) and \(C - B\)
Calculating \(B - A\):

\[ B - A = \frac{1}{\frac{1}{\sqrt{x}} + \frac{1}{\sqrt{z}}} - \frac{1}{\frac{1}{\sqrt{y}} + \frac{1}{\sqrt{z}}} \]

Finding a common denominator:

\[ = \frac{\left(\frac{1}{\sqrt{y}} + \frac{1}{\sqrt{z}}\right) - \left(\frac{1}{\sqrt{x}} + \frac{1}{\sqrt{z}}\right)}{\left(\frac{1}{\sqrt{x}} + \frac{1}{\sqrt{z}}\right)\left(\frac{1}{\sqrt{y}} + \frac{1}{\sqrt{z}}\right)} \]

This simplifies to:

\[ = \frac{\frac{1}{\sqrt{y}} - \frac{1}{\sqrt{x}}}{\left(\frac{1}{\sqrt{x}} + \frac{1}{\sqrt{z}}\right)\left(\frac{1}{\sqrt{y}} + \frac{1}{\sqrt{z}}\right)} \]

Now calculating \(C - B\):

\[ C - B = \frac{1}{\frac{1}{\sqrt{y}} + \frac{1}{\sqrt{x}}} - \frac{1}{\frac{1}{\sqrt{x}} + \frac{1}{\sqrt{z}}} \]

Following similar steps will yield:

\[ = \frac{\frac{1}{\sqrt{z}} - \frac{1}{\sqrt{y}}}{\left(\frac{1}{\sqrt{y}} + \frac{1}{\sqrt{x}}\right)\left(\frac{1}{\sqrt{x}} + \frac{1}{\sqrt{z}}\right)} \]

### Step 5: Establishing the Equality
From the property of harmonic progression, we know that:

\[ \frac{1}{y} - \frac{1}{x} = \frac{1}{z} - \frac{1}{y} \]

This implies:

\[ B - A = C - B \]

Thus, we have shown that:

\[ 2B = A + C \]

### Conclusion
Therefore, \(A\), \(B\), and \(C\) are in arithmetic progression.