If the sum of the roots of the quadratic equation `ax^(2) + bx + c = 0 ` is equal to the sum of the squares of their reciprocals, then a/c, b/a and c/b are in

Options

A. A.P
B. G.P
C. H.P
D. A.G.P

Answer

To solve the problem, we need to analyze the relationship between the roots of the quadratic equation \( ax^2 + bx + c = 0 \) and the conditions given in the question.

### Step-by-Step Solution:

1. **Identify the Roots**:
Let the roots of the quadratic equation be \( \alpha \) and \( \beta \).

2. **Sum of the Roots**:
From Vieta's formulas, we know:
\[ \alpha + \beta = -\frac{b}{a} \]

3. **Sum of the Squares of the Reciprocals**:
The sum of the squares of the reciprocals of the roots is given by:
\[ \frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\beta^2 + \alpha^2}{\alpha^2 \beta^2} \]
Using the identity \( \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \), we can express this as:
\[ \frac{(\alpha + \beta)^2 - 2\alpha\beta}{\alpha^2 \beta^2} \]

4. **Substituting Values**:
We know:
\[ \alpha \beta = \frac{c}{a} \]
Therefore, \( \alpha^2 \beta^2 = \left(\frac{c}{a}\right)^2 \).

Substituting these into the equation gives:
\[ \frac{(-\frac{b}{a})^2 - 2 \cdot \frac{c}{a}}{\left(\frac{c}{a}\right)^2} = \frac{\frac{b^2}{a^2} - \frac{2c}{a}}{\frac{c^2}{a^2}} = \frac{b^2 - 2ac}{c^2} \]

5. **Setting the Equation**:
According to the problem, we have:
\[ -\frac{b}{a} = \frac{b^2 - 2ac}{c^2} \]

6. **Cross Multiplying**:
Cross-multiplying gives:
\[ -bc^2 = a(b^2 - 2ac) \]

7. **Rearranging**:
Rearranging this equation leads to:
\[ ab^2 - 2ac^2 + bc^2 = 0 \]

8. **Dividing by \( abc \)**:
Dividing the entire equation by \( abc \) gives:
\[ \frac{b^2}{a} - \frac{2c}{b} + \frac{c}{a} = 0 \]

9. **Identifying the Sequence**:
Rearranging gives:
\[ \frac{b^2}{a} + \frac{c}{a} = \frac{2c}{b} \]
This implies that \( \frac{b}{c}, \frac{c}{a}, \frac{a}{b} \) are in Arithmetic Progression (AP).

10. **Conclusion**:
Since the problem asks for the ratios \( \frac{a}{c}, \frac{b}{a}, \frac{c}{b} \), and we found that their reciprocals are in AP, we conclude that:
\[ \frac{c}{b}, \frac{a}{c}, \frac{b}{a} \text{ are in Harmonic Progression (HP)}. \]

### Final Answer:
The values \( \frac{a}{c}, \frac{b}{a}, \frac{c}{b} \) are in Harmonic Progression (HP).