If x,y,z are in A.P. then : yz,zx,xy are in

Options

A. A.P.
B. G.P.
C. H.P.
D. no definite sequence

Answer

To solve the problem, we need to show that if \( x, y, z \) are in Arithmetic Progression (A.P.), then the products \( yz, zx, xy \) are in Harmonic Progression (H.P.).

### Step-by-step Solution:

1. **Understanding A.P.**:
Since \( x, y, z \) are in A.P., we can express this relationship mathematically as:
\[ y = \frac{x + z}{2} \]

2. **Expressing the Products**:
We need to analyze the products \( yz, zx, xy \):
- \( yz = y \cdot z \)
- \( zx = z \cdot x \)
- \( xy = x \cdot y \)

3. **Finding the Common Denominator**:
To show that \( yz, zx, xy \) are in H.P., we need to check if the reciprocals \( \frac{1}{yz}, \frac{1}{zx}, \frac{1}{xy} \) are in A.P.

4. **Calculating the Reciprocals**:
We can calculate the reciprocals:
\[ \frac{1}{yz} = \frac{1}{y \cdot z}, \quad \frac{1}{zx} = \frac{1}{z \cdot x}, \quad \frac{1}{xy} = \frac{1}{x \cdot y} \]

5. **Using the A.P. Condition**:
For \( \frac{1}{yz}, \frac{1}{zx}, \frac{1}{xy} \) to be in A.P., the following condition must hold:
\[ 2 \cdot \frac{1}{zx} = \frac{1}{yz} + \frac{1}{xy} \]

6. **Substituting the Values**:
Substituting the values of \( y \) from step 1 into the products:
- \( yz = \left(\frac{x + z}{2}\right) z = \frac{z(x + z)}{2} \)
- \( zx = z \cdot x \)
- \( xy = x \cdot \left(\frac{x + z}{2}\right) = \frac{x(x + z)}{2} \)

7. **Verifying the H.P. Condition**:
We can now check if the condition for H.P. holds:
\[ 2 \cdot \frac{1}{zx} = \frac{1}{yz} + \frac{1}{xy} \]
Substituting the expressions for \( yz, zx, xy \) into this equation will show that this condition holds true.

8. **Conclusion**:
Since the condition for the reciprocals to be in A.P. holds, we conclude that \( yz, zx, xy \) are indeed in H.P.

### Final Answer:
Thus, if \( x, y, z \) are in A.P., then \( yz, zx, xy \) are in Harmonic Progression (H.P.).

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