The sum of an infinite GP is x and the common ratio r is such that `|r|lt1`. If the first term of the GP is 2, then which one of the following is correct ?

Options

A. `-1ltxlt1`
B. `-ooltxlt1`
C. `1ltxltoo`
D. None of these

Explanation

To solve the problem, we need to find the relationship between the sum of an infinite geometric progression (GP) and its parameters. Given that the first term \( a = 2 \) and the sum \( S \) of the infinite GP is \( x \), we can use the formula for the sum of an infinite GP.

### Step-by-step Solution:

1. **Understand the formula for the sum of an infinite GP**:
The sum \( S \) of an infinite GP with first term \( a \) and common ratio \( r \) (where \( |r| < 1 \)) is given by:
\[ S = \frac{a}{1 - r} \]

2. **Substitute the known values**:
Here, we know that \( a = 2 \) and \( S = x \). Therefore, we can write:
\[ x = \frac{2}{1 - r} \]

3. **Rearranging the equation**:
To express \( r \) in terms of \( x \), we can rearrange the equation:
\[ 1 - r = \frac{2}{x} \]
From this, we can isolate \( r \):
\[ r = 1 - \frac{2}{x} \]

4. **Determine the constraints on \( r \)**:
Since we know that \( |r| < 1 \), we can set up the following inequalities:
\[ -1 < r < 1 \]

5. **Substituting for \( r \)**:
Substitute \( r = 1 - \frac{2}{x} \) into the inequalities:
\[ -1 < 1 - \frac{2}{x} < 1 \]

6. **Solving the left inequality**:
For the left part of the inequality:
\[ -1 < 1 - \frac{2}{x} \]
Rearranging gives:
\[ -2 < -\frac{2}{x} \implies 2 < \frac{2}{x} \implies x < 1 \]

7. **Solving the right inequality**:
For the right part of the inequality:
\[ 1 - \frac{2}{x} < 1 \]
This simplifies to:
\[ -\frac{2}{x} < 0 \implies x > 0 \]

8. **Combining the results**:
From the inequalities we derived, we have:
\[ 0 < x < 1 \]

### Conclusion:
The correct option based on the derived inequalities is that \( x \) must be greater than 2 and less than infinity, which means the correct answer is that \( x \) lies in the interval \( (2, \infty) \).