Three positive numbers form an increasing GP. If the middle terms in this GP is doubled, the new numbers are in AP. Then, the common ratio of the GP is

Options

A. `2-sqrt(3)`
B. `2+sqrt(3)`
C. `sqrt(2)+sqrt(3)`
D. `3+sqrt(2)`

Explanation

To solve the problem, we need to find the common ratio \( R \) of the geometric progression (GP) given that the middle term is doubled and the new terms form an arithmetic progression (AP).

### Step-by-Step Solution:

1. **Define the Terms of the GP**:
Let the three terms of the GP be:
\[ A, \quad AR, \quad AR^2 \]
where \( A \) is the first term and \( R \) is the common ratio.

2. **Double the Middle Term**:
If we double the middle term \( AR \), the new terms become:
\[ A, \quad 2AR, \quad AR^2 \]

3. **Condition for AP**:
The terms \( A, 2AR, AR^2 \) are in AP. For three numbers to be in AP, the following condition must hold:
\[ 2AR - A = AR^2 - 2AR \]
This simplifies to:
\[ 2AR - A = AR^2 - 2AR \]

4. **Rearranging the Equation**:
Rearranging gives:
\[ 2AR + 2AR = AR^2 + A \]
which simplifies to:
\[ 4AR = AR^2 + A \]

5. **Factoring Out A**:
Assuming \( A \neq 0 \), we can divide the entire equation by \( A \):
\[ 4R = R^2 + 1 \]

6. **Rearranging to Form a Quadratic Equation**:
Rearranging gives us:
\[ R^2 - 4R + 1 = 0 \]

7. **Using the Quadratic Formula**:
We can solve for \( R \) using the quadratic formula:
\[ R = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = -4, c = 1 \):
\[ R = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} \]
This simplifies to:
\[ R = \frac{4 \pm \sqrt{16 - 4}}{2} = \frac{4 \pm \sqrt{12}}{2} = \frac{4 \pm 2\sqrt{3}}{2} = 2 \pm \sqrt{3} \]

8. **Selecting the Positive Solution**:
Since \( R \) must be greater than 1 (as the GP is increasing), we choose:
\[ R = 2 + \sqrt{3} \]

### Final Answer:
The common ratio \( R \) of the GP is:
\[ \boxed{2 + \sqrt{3}} \]