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(3) - 2 `
D. `3 + sqrt(2)`

Answer

To solve the problem step by step, we will follow the reasoning used in the video transcript.

### Step-by-Step Solution:

1. **Define the Terms of the GP:**
Let the three positive numbers in the increasing geometric progression (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:**
According to the problem, if we double the middle term \( ar \), the new terms become:
\[ a, \quad 2ar, \quad ar^2 \]

3. **Condition for AP:**
The new terms \( a, 2ar, ar^2 \) are said to be in arithmetic progression (AP). For three numbers to be in AP, the middle term must be the average of the other two. Therefore, we have:
\[ 2ar - a = ar^2 - 2ar \]

4. **Simplifying the Equation:**
Rearranging the equation gives:
\[ 2ar - a = ar^2 - 2ar \]
This can be rewritten as:
\[ 2ar - a + 2ar = ar^2 \]
Simplifying further:
\[ 4ar - a = ar^2 \]

5. **Rearranging the Equation:**
We can rearrange this to:
\[ ar^2 - 4ar + a = 0 \]

6. **Dividing by \( a \):**
Since \( a \) is positive, we can divide the entire equation by \( a \):
\[ r^2 - 4r + 1 = 0 \]

7. **Using the Quadratic Formula:**
Now we can apply the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = -4, c = 1 \):
\[ r = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} \]
Simplifying this gives:
\[ r = \frac{4 \pm \sqrt{16 - 4}}{2} = \frac{4 \pm \sqrt{12}}{2} \]
\[ r = \frac{4 \pm 2\sqrt{3}}{2} = 2 \pm \sqrt{3} \]

8. **Selecting the Valid Solution:**
Since the problem states that the numbers are positive and in increasing order, we need \( r > 1 \). Therefore, we choose:
\[ r = 2 + \sqrt{3} \]

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