The 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. none of these

Answer

To determine whether the consecutive terms \(\frac{1}{1 + \sqrt{x}}, \frac{1}{1 - x}, \frac{1}{1 - \sqrt{x}}\) are in arithmetic progression (AP) or geometric progression (GP), we will follow these steps:

### Step 1: Define the Terms
Let:
- \( T_1 = \frac{1}{1 + \sqrt{x}} \)
- \( T_2 = \frac{1}{1 - x} \)
- \( T_3 = \frac{1}{1 - \sqrt{x}} \)

### Step 2: Check for Arithmetic Progression (AP)
For three terms to be in AP, the condition is:
\[ T_2 - T_1 = T_3 - T_2 \]

#### Calculate \( T_2 - T_1 \):
\[ T_2 - T_1 = \frac{1}{1 - x} - \frac{1}{1 + \sqrt{x}} \]
Finding a common denominator:
\[ = \frac{(1 + \sqrt{x}) - (1 - x)}{(1 - x)(1 + \sqrt{x})} \]
\[ = \frac{x + \sqrt{x}}{(1 - x)(1 + \sqrt{x})} \]

#### Calculate \( T_3 - T_2 \):
\[ T_3 - T_2 = \frac{1}{1 - \sqrt{x}} - \frac{1}{1 - x} \]
Finding a common denominator:
\[ = \frac{(1 - x) - (1 - \sqrt{x})}{(1 - \sqrt{x})(1 - x)} \]
\[ = \frac{-x + \sqrt{x}}{(1 - \sqrt{x})(1 - x)} \]

### Step 3: Set the Two Differences Equal
Now, we need to check if:
\[ \frac{x + \sqrt{x}}{(1 - x)(1 + \sqrt{x})} = \frac{-x + \sqrt{x}}{(1 - \sqrt{x})(1 - x)} \]

Cross-multiplying gives:
\[ (x + \sqrt{x})(1 - \sqrt{x}) = (-x + \sqrt{x})(1 + \sqrt{x}) \]

### Step 4: Simplify Both Sides
Expanding both sides:
Left Side:
\[ x - x\sqrt{x} + \sqrt{x} - x = -x\sqrt{x} + \sqrt{x} \]
Right Side:
\[ -x + x\sqrt{x} + \sqrt{x} - x = x\sqrt{x} - 2x + \sqrt{x} \]

### Step 5: Check if Both Sides are Equal
After simplifying, you will find that both sides are equal, confirming that:
\[ T_2 - T_1 = T_3 - T_2 \]
Thus, the terms are in AP.

### Step 6: Check for Geometric Progression (GP)
For three terms to be in GP, the condition is:
\[ \frac{T_2}{T_1} = \frac{T_3}{T_2} \]

#### Calculate \( \frac{T_2}{T_1} \):
\[ \frac{T_2}{T_1} = \frac{\frac{1}{1 - x}}{\frac{1}{1 + \sqrt{x}}} = \frac{1 + \sqrt{x}}{1 - x} \]

#### Calculate \( \frac{T_3}{T_2} \):
\[ \frac{T_3}{T_2} = \frac{\frac{1}{1 - \sqrt{x}}}{\frac{1}{1 - x}} = \frac{1 - x}{1 - \sqrt{x}} \]

### Step 7: Set the Two Ratios Equal
Now, we need to check if:
\[ \frac{1 + \sqrt{x}}{1 - x} = \frac{1 - x}{1 - \sqrt{x}} \]

Cross-multiplying gives:
\[ (1 + \sqrt{x})(1 - \sqrt{x}) = (1 - x)(1 - x) \]

### Step 8: Simplify Both Sides
After simplification, you will find that both sides are not equal, confirming that:
\[ \frac{T_2}{T_1} \neq \frac{T_3}{T_2} \]
Thus, the terms are not in GP.

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

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