If a, b, c form a G.P. with common ratio r such that 0 < r < 1, and if `a,3/2b ,-4c` form an AP, then r is equal to

Options

A. `(1)/(2)`
B. `(2)/(3)`
C. `(1)/(3)`
D. `(1)/(4)`

Answer

To solve the problem, we need to find the common ratio \( r \) of the geometric progression (G.P.) formed by \( a, b, c \) under the condition that \( a, \frac{3}{2}b, -4c \) form an arithmetic progression (A.P.).

### Step-by-Step Solution:

1. **Understanding the G.P.**:
Since \( a, b, c \) are in G.P. with common ratio \( r \), we can express \( b \) and \( c \) in terms of \( a \):
\[ b = ar \quad \text{and} \quad c = ar^2 \]

2. **Setting up the A.P. condition**:
The terms \( a, \frac{3}{2}b, -4c \) form an A.P. This means that:
\[ 2 \cdot \frac{3}{2}b = a + (-4c) \]
Simplifying this gives:
\[ 3b = a - 4c \]

3. **Substituting the values of \( b \) and \( c \)**:
Substitute \( b = ar \) and \( c = ar^2 \) into the equation:
\[ 3(ar) = a - 4(ar^2) \]
This simplifies to:
\[ 3ar = a - 4ar^2 \]

4. **Rearranging the equation**:
Rearranging the equation gives:
\[ 4ar^2 + 3ar - a = 0 \]
Factoring out \( a \) (assuming \( a \neq 0 \)):
\[ a(4r^2 + 3r - 1) = 0 \]
Thus, we need to solve:
\[ 4r^2 + 3r - 1 = 0 \]

5. **Using the quadratic formula**:
The quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) can be applied here, where \( a = 4, b = 3, c = -1 \):
\[ r = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 4 \cdot (-1)}}{2 \cdot 4} \]
Simplifying the discriminant:
\[ r = \frac{-3 \pm \sqrt{9 + 16}}{8} = \frac{-3 \pm \sqrt{25}}{8} = \frac{-3 \pm 5}{8} \]

6. **Calculating the roots**:
This gives us two potential solutions:
\[ r = \frac{2}{8} = \frac{1}{4} \quad \text{and} \quad r = \frac{-8}{8} = -1 \]

7. **Considering the range of \( r \)**:
Since it is given that \( 0 < r < 1 \), we discard \( r = -1 \) and keep:
\[ r = \frac{1}{4} \]

### Final Answer:
Thus, the common ratio \( r \) is:
\[ \boxed{\frac{1}{4}} \]