The following consecutive terms `(1)/(1+sqrt(x)),(1)/(1-x),(1)/(1-sqrt(x))` of a series are in

Options

A. H.P.
B. G.P
C. A.P.
D. A.P., G.P.

Answer

To determine the relationship between the consecutive terms \( \frac{1}{1+\sqrt{x}}, \frac{1}{1-x}, \frac{1}{1-\sqrt{x}} \), we need to check if they are in Arithmetic Progression (AP), Geometric Progression (GP), or Harmonic Progression (HP).

### Step 1: Define the terms
Let:
- \( a = \frac{1}{1+\sqrt{x}} \)
- \( b = \frac{1}{1-x} \)
- \( c = \frac{1}{1-\sqrt{x}} \)

### Step 2: Check for Arithmetic Progression (AP)
For the terms to be in AP, the following condition must hold:
\[ 2b = a + c \]
Substituting the values of \( a \), \( b \), and \( c \):
\[ 2 \cdot \frac{1}{1-x} = \frac{1}{1+\sqrt{x}} + \frac{1}{1-\sqrt{x}} \]

### Step 3: Simplify the right-hand side
To simplify \( \frac{1}{1+\sqrt{x}} + \frac{1}{1-\sqrt{x}} \), we find a common denominator:
\[ \frac{1}{1+\sqrt{x}} + \frac{1}{1-\sqrt{x}} = \frac{(1-\sqrt{x}) + (1+\sqrt{x})}{(1+\sqrt{x})(1-\sqrt{x})} \]
This simplifies to:
\[ \frac{2}{1-x} \]
since \( (1+\sqrt{x})(1-\sqrt{x}) = 1 - x \).

### Step 4: Substitute back into the AP condition
Now substituting back into the AP condition:
\[ 2 \cdot \frac{1}{1-x} = \frac{2}{1-x} \]
This equation holds true, confirming that the terms are in AP.

### Conclusion
Thus, the consecutive terms \( \frac{1}{1+\sqrt{x}}, \frac{1}{1-x}, \frac{1}{1-\sqrt{x}} \) are in Arithmetic Progression (AP).