Let a,b,c are in G.P. with common ratio `r(0ltrlt1)`. Then a,2b,3c are in A.P., if r rquals:

Options

A. 44228
B. `1/sqrt2`
C. 44256
D. none

Answer

To solve the problem, we need to find the value of the common ratio \( r \) given that \( a, b, c \) are in a geometric progression (G.P.) and \( a, 2b, 3c \) are in an arithmetic progression (A.P.).

### Step-by-Step Solution:

1. **Define the terms of the G.P.**:
Since \( a, b, c \) are in G.P. with common ratio \( r \), we can express:
- \( b = ar \)
- \( c = ar^2 \)

2. **Set up the condition for A.P.**:
For \( a, 2b, 3c \) to be in A.P., the condition is:
\[ 2b = a + 3c \]

3. **Substitute the values of \( b \) and \( c \)**:
Substitute \( b \) and \( c \) from the G.P. definitions into the A.P. condition:
\[ 2(ar) = a + 3(ar^2) \]
Simplifying this gives:
\[ 2ar = a + 3ar^2 \]

4. **Rearrange the equation**:
Move all terms to one side:
\[ 3ar^2 - 2ar + a = 0 \]

5. **Factor out \( a \)**:
Since \( a \neq 0 \) (as it is a term in G.P.), we can factor out \( a \):
\[ a(3r^2 - 2r + 1) = 0 \]
This leads us to solve the quadratic equation:
\[ 3r^2 - 2r + 1 = 0 \]

6. **Use the quadratic formula**:
The quadratic formula is given by:
\[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 3, b = -2, c = 1 \):
\[ r = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 3 \cdot 1}}{2 \cdot 3} \]
\[ r = \frac{2 \pm \sqrt{4 - 12}}{6} \]
\[ r = \frac{2 \pm \sqrt{-8}}{6} \]
Since the discriminant is negative, we have complex roots. However, we can also check if we made a mistake in our equation setup.

7. **Revisit the equation**:
We should have:
\[ 3r^2 - 4r + 1 = 0 \]
Using the quadratic formula again:
\[ r = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 3 \cdot 1}}{2 \cdot 3} \]
\[ r = \frac{4 \pm \sqrt{16 - 12}}{6} \]
\[ r = \frac{4 \pm \sqrt{4}}{6} \]
\[ r = \frac{4 \pm 2}{6} \]
This gives us two possible values:
\[ r = \frac{6}{6} = 1 \quad \text{and} \quad r = \frac{2}{6} = \frac{1}{3} \]

8. **Select the valid solution**:
Since \( r \) must be between 0 and 1, we discard \( r = 1 \) and accept:
\[ r = \frac{1}{3} \]

### Conclusion:
The value of \( r \) is \( \frac{1}{3} \).