If a,b, c are in G.P., then the equations `ax^(2) + 2bx + c = 0 and dx^(2) + 2ex + f = 0` have common root if `(d)/(a), (e)/(b), (f)/(c)` are in

Options

A. A.P.
B. G.P.
C. H.P.
D. None of these

Explanation

To solve the problem, we need to establish the relationship between the coefficients of the two quadratic equations given that \( a, b, c \) are in geometric progression (G.P.) and that the two equations have a common root.

### Step-by-Step Solution:

1. **Understanding G.P. Condition**:
Since \( a, b, c \) are in G.P., we have the condition:
\[ b^2 = ac \]

2. **Common Root**:
Let \( r \) be the common root of the equations \( ax^2 + 2bx + c = 0 \) and \( dx^2 + 2ex + f = 0 \). Therefore, substituting \( r \) into both equations gives:
\[ ar^2 + 2br + c = 0 \quad \text{(1)} \]
\[ dr^2 + 2er + f = 0 \quad \text{(2)} \]

3. **Expressing \( r^2 \) from Equation (1)**:
From equation (1), we can express \( r^2 \):
\[ ar^2 = -2br - c \implies r^2 = \frac{-2br - c}{a} \quad \text{(3)} \]

4. **Substituting \( r^2 \) in Equation (2)**:
Substitute equation (3) into equation (2):
\[ d\left(\frac{-2br - c}{a}\right) + 2er + f = 0 \]
Multiplying through by \( a \) to eliminate the fraction:
\[ d(-2br - c) + 2ear + fa = 0 \]
Rearranging gives:
\[ -2dbr - dc + 2ear + fa = 0 \]

5. **Grouping Terms**:
Group the terms involving \( r \):
\[ (-2db + 2ea)r + (fa - dc) = 0 \]
For this equation to hold for all \( r \), both coefficients must equal zero:
\[ -2db + 2ea = 0 \quad \text{(4)} \]
\[ fa - dc = 0 \quad \text{(5)} \]

6. **Solving Equation (4)**:
From equation (4):
\[ 2ea = 2db \implies \frac{d}{a} = \frac{e}{b} \quad \text{(6)} \]

7. **Solving Equation (5)**:
From equation (5):
\[ fa = dc \implies \frac{f}{c} = \frac{d}{a} \quad \text{(7)} \]

8. **Conclusion**:
From equations (6) and (7), we have:
\[ \frac{d}{a}, \frac{e}{b}, \frac{f}{c} \text{ are in A.P.} \]

### Final Answer:
Thus, if \( a, b, c \) are in G.P., then the ratios \( \frac{d}{a}, \frac{e}{b}, \frac{f}{c} \) are in Arithmetic Progression (A.P.).

---