If x, y, z be in A.P., then `x + (1)/(yz), y + (1)/(zx), z + (1)/(xy)` are in

Options

A. A.P.
B. G.P.
C. H.P.
D. none

Answer

To solve the problem, we need to show that if \( x, y, z \) are in arithmetic progression (A.P.), then the expressions \( x + \frac{1}{yz}, y + \frac{1}{zx}, z + \frac{1}{xy} \) are also in A.P.

### Step-by-Step Solution:

1. **Understanding A.P. Condition**:
Since \( x, y, z \) are in A.P., we have:
\[ y - x = z - y \]
This implies:
\[ 2y = x + z \quad \text{(1)} \]

2. **Finding the First Difference**:
We need to check if the difference between the first two terms is equal to the difference between the last two terms.
Let's denote:
\[ a_1 = x + \frac{1}{yz}, \quad a_2 = y + \frac{1}{zx} \]
We calculate \( a_2 - a_1 \):
\[ a_2 - a_1 = \left( y + \frac{1}{zx} \right) - \left( x + \frac{1}{yz} \right) \]
Simplifying this gives:
\[ a_2 - a_1 = (y - x) + \left( \frac{1}{zx} - \frac{1}{yz} \right) \]
The second part can be simplified:
\[ \frac{1}{zx} - \frac{1}{yz} = \frac{y - x}{zyx} \]
Thus:
\[ a_2 - a_1 = (y - x) + \frac{y - x}{zyx} = (y - x) \left( 1 + \frac{1}{zyx} \right) \quad \text{(2)} \]

3. **Finding the Second Difference**:
Now, we calculate the difference between the last two terms:
\[ a_3 = z + \frac{1}{xy} \]
So we calculate \( a_3 - a_2 \):
\[ a_3 - a_2 = \left( z + \frac{1}{xy} \right) - \left( y + \frac{1}{zx} \right) \]
Simplifying this gives:
\[ a_3 - a_2 = (z - y) + \left( \frac{1}{xy} - \frac{1}{zx} \right) \]
The second part can be simplified:
\[ \frac{1}{xy} - \frac{1}{zx} = \frac{z - y}{xyz} \]
Thus:
\[ a_3 - a_2 = (z - y) + \frac{z - y}{xyz} = (z - y) \left( 1 + \frac{1}{xyz} \right) \quad \text{(3)} \]

4. **Comparing the Differences**:
From equations (2) and (3), we need to check if:
\[ (y - x) \left( 1 + \frac{1}{zyx} \right) = (z - y) \left( 1 + \frac{1}{xyz} \right) \]
Since \( y - x = z - y \) (from the A.P. condition), we can substitute \( z - y \) with \( y - x \). Thus, both expressions will be equal, confirming that:
\[ a_2 - a_1 = a_3 - a_2 \]

5. **Conclusion**:
Since the common difference between the first two terms is equal to the common difference between the last two terms, we conclude that:
\[ x + \frac{1}{yz}, y + \frac{1}{zx}, z + \frac{1}{xy} \text{ are in A.P.} \]