If x is a positive real number different from 1, then prove that the numbers `1/(1+sqrtx),1/(1-x),1/(1-sqrtx),…` are in A.P. Also find their common difference.

Explanation

To prove that the numbers \( \frac{1}{1+\sqrt{x}}, \frac{1}{1-x}, \frac{1}{1-\sqrt{x}} \) are in Arithmetic Progression (A.P.), we need to show that the common difference between consecutive terms is the same.

### Step 1: Identify the terms
Let:
- \( a_1 = \frac{1}{1+\sqrt{x}} \)
- \( a_2 = \frac{1}{1-x} \)
- \( a_3 = \frac{1}{1-\sqrt{x}} \)

### Step 2: Calculate the common difference \( d \)
The common difference \( d \) can be calculated as:
\[ d = a_2 - a_1 = \frac{1}{1-x} - \frac{1}{1+\sqrt{x}} \]

### Step 3: Find a common denominator
To subtract these fractions, we need to find a common denominator:
\[ d = \frac{(1+\sqrt{x}) - (1-x)}{(1-x)(1+\sqrt{x})} \]

### Step 4: Simplify the numerator
Now simplify the numerator:
\[ (1+\sqrt{x}) - (1-x) = \sqrt{x} + x \]
Thus, we have:
\[ d = \frac{\sqrt{x} + x}{(1-x)(1+\sqrt{x})} \]

### Step 5: Factor out \( \sqrt{x} \)
We can factor \( \sqrt{x} \) from the numerator:
\[ d = \frac{\sqrt{x}(1+\sqrt{x})}{(1-x)(1+\sqrt{x})} \]

### Step 6: Cancel out \( (1+\sqrt{x}) \)
Since \( 1+\sqrt{x} \) is not zero (as \( x \) is a positive real number), we can cancel it out:
\[ d = \frac{\sqrt{x}}{1-x} \]

### Step 7: Check the second common difference
Now we need to check if \( d = a_3 - a_2 \):
\[ d = a_3 - a_2 = \frac{1}{1-\sqrt{x}} - \frac{1}{1-x} \]

### Step 8: Find a common denominator for the second difference
Using a common denominator:
\[ d = \frac{(1-x) - (1-\sqrt{x})}{(1-\sqrt{x})(1-x)} \]

### Step 9: Simplify the numerator
The numerator simplifies to:
\[ (1-x) - (1-\sqrt{x}) = \sqrt{x} - x \]
Thus, we have:
\[ d = \frac{\sqrt{x} - x}{(1-\sqrt{x})(1-x)} \]

### Step 10: Factor out \( \sqrt{x} \)
We can factor \( \sqrt{x} \) from the numerator:
\[ d = \frac{\sqrt{x}(1 - \sqrt{x})}{(1-\sqrt{x})(1-x)} \]

### Step 11: Cancel out \( (1-\sqrt{x}) \)
Since \( 1-\sqrt{x} \) is not zero (as \( x \) is a positive real number different from 1), we can cancel it out:
\[ d = \frac{\sqrt{x}}{1-x} \]

### Conclusion
Since both calculations of the common difference yield the same result:
\[ d = \frac{\sqrt{x}}{1-x} \]
we conclude that the numbers \( \frac{1}{1+\sqrt{x}}, \frac{1}{1-x}, \frac{1}{1-\sqrt{x}} \) are in A.P. with a common difference of \( \frac{\sqrt{x}}{1-x} \).