If the equations `ax^(2) +2bx +c = 0 and Ax^(2) +2Bx+C=0` have a common root and a,b,c are in G.P prove that `a/A , b/B, c/C ` are in H.P

Answer

To solve the problem, we need to prove that if the equations \( ax^2 + 2bx + c = 0 \) and \( Ax^2 + 2Bx + C = 0 \) have a common root and \( a, b, c \) are in G.P., then \( \frac{a}{A}, \frac{b}{B}, \frac{c}{C} \) are in H.P.

### Step-by-Step Solution:

1. **Understanding the Given Information**:
We know that \( a, b, c \) are in G.P. This implies that:
\[ b^2 = ac \]

2. **Common Root**:
Let the common root of the equations be \( r \). Then both equations can be expressed as:
\[ ar^2 + 2br + c = 0 \quad \text{(1)} \]
\[ Ar^2 + 2Br + C = 0 \quad \text{(2)} \]

3. **Substituting the Common Root**:
From equation (1):
\[ ar^2 + 2br + c = 0 \implies c = -ar^2 - 2br \quad \text{(3)} \]
From equation (2):
\[ Ar^2 + 2Br + C = 0 \implies C = -Ar^2 - 2Br \quad \text{(4)} \]

4. **Using the G.P. Condition**:
Substitute \( c \) from (3) into the G.P. condition:
\[ b^2 = a(-ar^2 - 2br) \]
This simplifies to:
\[ b^2 = -a^2r^2 - 2abr \]

5. **Rearranging the Terms**:
Rearranging gives:
\[ a^2r^2 + 2abr + b^2 = 0 \]
This is a quadratic in \( r \).

6. **Finding the Roots**:
The discriminant of this quadratic must be non-negative for \( r \) to be real:
\[ D = (2ab)^2 - 4a^2b^2 = 0 \]
This indicates that there is exactly one solution for \( r \).

7. **Establishing the Relationship**:
Since both equations have the same root, we can equate the coefficients of the corresponding powers of \( x \) from both equations.

8. **Using the Coefficients**:
From the common root condition, we can derive:
\[ \frac{A}{a} = \frac{C}{c} = \frac{B}{b} \quad \text{(5)} \]

9. **Proving H.P.**:
From (5), we can express:
\[ \frac{a}{A}, \frac{b}{B}, \frac{c}{C} \]
If we denote:
\[ \frac{a}{A} = x, \quad \frac{b}{B} = y, \quad \frac{c}{C} = z \]
Then, \( x, y, z \) are in H.P. if:
\[ \frac{1}{x}, \frac{1}{y}, \frac{1}{z} \text{ are in A.P.} \]

10. **Conclusion**:
Since \( \frac{A}{a}, \frac{B}{b}, \frac{C}{c} \) are in A.P., it follows that \( \frac{a}{A}, \frac{b}{B}, \frac{c}{C} \) are in H.P.