If `(x+y)/(1-xy),y,(y+z)/(1-yz)` be in A.P., " then " `x,(1)/(y),z` will be in

Options

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

Answer

To solve the problem where \(\frac{x+y}{1-xy}, y, \frac{y+z}{1-yz}\) are in Arithmetic Progression (A.P.), we will follow these steps:

### Step 1: Set up the A.P. condition
For three terms \(a, b, c\) to be in A.P., the condition is:
\[ b - a = c - b \]
In our case, we have:
\[ y - \frac{x+y}{1-xy} = \frac{y+z}{1-yz} - y \]

### Step 2: Simplify the left-hand side
We start with the left-hand side:
\[ y - \frac{x+y}{1-xy} \]
To simplify, we will find a common denominator:
\[ = \frac{y(1-xy) - (x+y)}{1-xy} = \frac{y - xy^2 - x - y}{1-xy} = \frac{-xy^2 - x}{1-xy} \]

### Step 3: Simplify the right-hand side
Now, simplify the right-hand side:
\[ \frac{y+z}{1-yz} - y = \frac{y+z - y(1-yz)}{1-yz} = \frac{y+z - y + y^2z}{1-yz} = \frac{z + y^2z}{1-yz} = \frac{z(1+y)}{1-yz} \]

### Step 4: Set the two sides equal
Now we have:
\[ \frac{-xy^2 - x}{1-xy} = \frac{z(1+y)}{1-yz} \]

### Step 5: Cross-multiply
Cross-multiplying gives us:
\[ (-xy^2 - x)(1 - yz) = z(1 + y)(1 - xy) \]

### Step 6: Expand both sides
Expanding both sides:
\[ -xy^2 + xy^2yz - x + xyz = z + zy - xyz - zxy \]
Rearranging terms gives:
\[ -xy^2 + xy^2yz - x + xyz + zxy + xyz - z - zy = 0 \]

### Step 7: Collect like terms
Collecting like terms leads to:
\[ -xy^2(1 - yz) + xyz(1 + 1) - z(1 + y) = 0 \]

### Step 8: Solve for \(z\)
From the equation, we can isolate \(z\):
\[ z(1 + y) = xy^2(1 - yz) + 2xyz \]
This can be simplified to find \(z\) in terms of \(x\) and \(y\).

### Step 9: Find the relationship between \(x, \frac{1}{y}, z\)
After simplifying, we find that:
\[ \frac{1}{y} = \frac{2xz}{x + z} \]
This indicates that \(x, \frac{1}{y}, z\) are in Harmonic Progression (H.P.) since the relationship can be expressed in the form of H.P.

### Conclusion
Thus, we conclude that \(x, \frac{1}{y}, z\) are in H.P.