The sum of the roots of the equation `ax^(2)+x+c=0`( where a and c are non-zero) is equal to the sum of the reciprocals of their squares. Then `a,ca^(2),c^(2)` are in

Options

A. AP
B. GP
C. HP
D. None of the above

Explanation

To solve the problem, we need to analyze the quadratic equation given and derive the required relationships.

### Step 1: Identify the roots and their properties
The quadratic equation is given by:
\[ ax^2 + x + c = 0 \]

Let the roots of this equation be \( \alpha \) and \( \beta \).

### Step 2: Use Vieta's formulas
From Vieta's formulas, we know:
- The sum of the roots \( \alpha + \beta = -\frac{b}{a} = -\frac{1}{a} \)
- The product of the roots \( \alpha \beta = \frac{c}{a} \)

### Step 3: Set up the equation for the sum of the reciprocals of their squares
We are given that the sum of the roots is equal to the sum of the reciprocals of their squares:
\[ \alpha + \beta = \frac{1}{\alpha^2} + \frac{1}{\beta^2} \]

### Step 4: Express the sum of the reciprocals of the squares
The sum of the reciprocals of the squares can be expressed as:
\[ \frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\beta^2 + \alpha^2}{\alpha^2 \beta^2} \]

Using the identity \( \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \):
\[ \alpha^2 + \beta^2 = \left(-\frac{1}{a}\right)^2 - 2\left(\frac{c}{a}\right) = \frac{1}{a^2} - \frac{2c}{a} \]

### Step 5: Substitute into the equation
Now substituting back, we have:
\[ \frac{\frac{1}{a^2} - \frac{2c}{a}}{\left(\frac{c}{a}\right)^2} = \frac{\frac{1}{a^2} - \frac{2c}{a}}{\frac{c^2}{a^2}} = \frac{a^2\left(\frac{1}{a^2} - \frac{2c}{a}\right)}{c^2} = \frac{1 - 2ac}{c^2} \]

### Step 6: Set the two sides equal
Now we set the two expressions equal to each other:
\[ -\frac{1}{a} = \frac{1 - 2ac}{c^2} \]

### Step 7: Cross-multiply and simplify
Cross-multiplying gives:
\[ -c^2 = a(1 - 2ac) \]
\[ -c^2 = a - 2a^2c \]
Rearranging gives:
\[ 2a^2c - a - c^2 = 0 \]

### Step 8: Analyze the relationship
This is a quadratic equation in terms of \( c \). The discriminant must be non-negative for \( c \) to have real values:
\[ D = (-1)^2 - 4 \cdot 2a^2 \cdot (-1) \]
\[ D = 1 + 8a^2 \]
Since \( D \) is always positive for \( a \neq 0 \), \( c \) can take real values.

### Conclusion
Now, we need to check the relationship between \( a, ca^2, c^2 \). Since \( c \) is expressed in terms of \( a \), we can conclude that:
- \( a, ca^2, c^2 \) are in a geometric progression.