Three numbers form an increasing G.P. . If the middle number is doubled, then the new numbers are in A.P. The common ratio of the G.P is:

Options

A. a)`2-sqrt3`
B. b)`2+sqrt3`
C. c)`sqrt3-2`
D. d)`3+sqrt2`

Answer

To solve the problem, we need to find the common ratio of three numbers that form an increasing geometric progression (G.P.) and, when the middle number is doubled, the new numbers form an arithmetic progression (A.P.).

### Step-by-Step Solution:

1. **Define the Terms of the G.P.**:
Let the three numbers in the G.P. be:
- First term: \( a \)
- Second term: \( ar \) (middle term)
- Third term: \( ar^2 \)

2. **Doubling the Middle Term**:
If we double the middle term, the new terms become:
- First term: \( a \)
- New middle term: \( 2ar \)
- Third term: \( ar^2 \)

3. **Condition for A.P.**:
For the three numbers \( a, 2ar, ar^2 \) to be in A.P., the condition is:
\[ 2ar - a = ar^2 - 2ar \]
This simplifies to:
\[ 2ar - a = ar^2 - 2ar \]

4. **Rearranging the Equation**:
Rearranging gives:
\[ 2ar - a = ar^2 - 2ar \]
Combine like terms:
\[ 4ar - a = ar^2 \]

5. **Factoring Out Common Terms**:
Factor out \( a \):
\[ a(4r - 1) = ar^2 \]
Assuming \( a \neq 0 \), we can divide both sides by \( a \):
\[ 4r - 1 = r^2 \]

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, \( a = 1, b = -4, c = 1 \):
\[ r = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} \]
\[ r = \frac{4 \pm \sqrt{16 - 4}}{2} \]
\[ r = \frac{4 \pm \sqrt{12}}{2} \]
\[ r = \frac{4 \pm 2\sqrt{3}}{2} \]
\[ r = 2 \pm \sqrt{3} \]

8. **Selecting the Positive Ratio**:
Since the G.P. is increasing, we take the positive value:
\[ r = 2 + \sqrt{3} \]

### Final Answer:
The common ratio of the G.P. is \( r = 2 + \sqrt{3} \).