Suppose a,b,c are distinct positive real numbers such that a,2b,3c are in A.P. and a,b,c are in G.P. The common ratio of G.P. is

Options

A. 2
B. `1//2`
C. `1//3`
D. 3

Answer

To solve the problem, we need to use the properties of Arithmetic Progression (A.P.) and Geometric Progression (G.P.) as given in the question.

### Step-by-Step Solution:

1. **Understanding the conditions**:
- We know that \( a, 2b, 3c \) are in A.P. This means:
\[ 2b = \frac{a + 3c}{2} \]
- We also know that \( a, b, c \) are in G.P. This means:
\[ b^2 = ac \]

2. **From the A.P. condition**:
- Rearranging the A.P. equation gives:
\[ 4b = a + 3c \quad \text{(1)} \]

3. **From the G.P. condition**:
- We can express \( c \) in terms of \( a \) and \( b \):
\[ c = \frac{b^2}{a} \quad \text{(2)} \]

4. **Substituting (2) into (1)**:
- Substitute \( c \) from equation (2) into equation (1):
\[ 4b = a + 3\left(\frac{b^2}{a}\right) \]
- This simplifies to:
\[ 4b = a + \frac{3b^2}{a} \]
- Multiplying through by \( a \) to eliminate the fraction:
\[ 4ab = a^2 + 3b^2 \quad \text{(3)} \]

5. **Rearranging equation (3)**:
- Rearranging gives us a quadratic equation in terms of \( a \):
\[ a^2 - 4ab + 3b^2 = 0 \]

6. **Using the quadratic formula**:
- The quadratic formula \( a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) can be applied here:
\[ a = \frac{4b \pm \sqrt{(4b)^2 - 4 \cdot 1 \cdot 3b^2}}{2 \cdot 1} \]
- This simplifies to:
\[ a = \frac{4b \pm \sqrt{16b^2 - 12b^2}}{2} = \frac{4b \pm 2b}{2} \]
- Thus, we have two possible values for \( a \):
\[ a = 3b \quad \text{or} \quad a = b \]

7. **Finding the common ratio**:
- Since \( a, b, c \) are distinct positive real numbers, we discard \( a = b \).
- Therefore, we have \( a = 3b \).
- Now substituting \( a = 3b \) back into the G.P. condition \( b^2 = ac \):
\[ b^2 = (3b)c \implies c = \frac{b^2}{3b} = \frac{b}{3} \]

8. **Finding the common ratio**:
- The common ratio \( r \) of the G.P. is given by:
\[ r = \frac{b}{a} = \frac{b}{3b} = \frac{1}{3} \]

### Final Answer:
The common ratio of the G.P. is \( \frac{1}{3} \).