If `a, a_1 ,a_2----a_(2n-1),b` are in `A.P and a,b_1,b_2-----b_(2n-1),b` are in `G.P and a,c_1,c_2----c_(2n-1),b` are in `H.P` (which are non-zero and a,b are positive real numbers), then the roots of the equation `a_nx^2-b_nx |c_n=0` are (a) real and unequal (b) real and equal (c) imaginary (d) none of these

Options

A. real and unequal
B. real and equal
C. imaginary
D. none of these

Answer

To solve the problem, we need to analyze the given sequences and their properties. Here’s the step-by-step solution:

### Step 1: Understanding the sequences
We have three sequences:
1. \( a, a_1, a_2, \ldots, a_{2n-1}, b \) in Arithmetic Progression (A.P)
2. \( a, b_1, b_2, \ldots, b_{2n-1}, b \) in Geometric Progression (G.P)
3. \( a, c_1, c_2, \ldots, c_{2n-1}, b \) in Harmonic Progression (H.P)

### Step 2: Finding the values of \( a_n, b_n, c_n \)
- For the A.P, the \( n \)-th term \( a_n \) is given by:
\[ a_n = \frac{a + b}{2} \]

- For the G.P, the \( n \)-th term \( b_n \) is given by:
\[ b_n = \sqrt{ab} \]

- For the H.P, the \( n \)-th term \( c_n \) is given by:
\[ c_n = \frac{2ab}{a + b} \]

### Step 3: Formulating the quadratic equation
The quadratic equation given is:
\[ a_n x^2 - b_n x + c_n = 0 \]
Substituting the values we found:
\[ \left(\frac{a + b}{2}\right)x^2 - \sqrt{ab} x + \frac{2ab}{a + b} = 0 \]

### Step 4: Finding the discriminant
The discriminant \( D \) of a quadratic equation \( Ax^2 + Bx + C = 0 \) is given by:
\[ D = B^2 - 4AC \]
Here, \( A = \frac{a + b}{2} \), \( B = -\sqrt{ab} \), and \( C = \frac{2ab}{a + b} \).

Calculating \( D \):
\[ D = (-\sqrt{ab})^2 - 4 \left(\frac{a + b}{2}\right) \left(\frac{2ab}{a + b}\right) \]
\[ D = ab - 4 \cdot \frac{(a + b)}{2} \cdot \frac{2ab}{a + b} \]
\[ D = ab - 4ab = -3ab \]

### Step 5: Analyzing the discriminant
Since \( a \) and \( b \) are positive real numbers, \( -3ab < 0 \). This means that the discriminant \( D \) is negative.

### Conclusion
When the discriminant of a quadratic equation is negative, the roots are imaginary.

Thus, the answer to the question is:
**(c) imaginary**