Given that `alpha,gamma` are roots of the equation `A x^2-4x+1=0,a n dbeta,delta` the roots of the equation of `B x^2-6x+1=0,` such that `alpha,beta,gamma,a n ddelta` are in H.P., then a.`A=3` b. `A=4` `B=2` d. `B=8`

Options

A. A = 3
B. A = 4
C. B = 2
D. B = 8

Explanation

To solve the problem, we need to analyze the given information about the roots of the quadratic equations and their relationship in Harmonic Progression (H.P.).

### Step 1: Identify the roots and their relationships
Given:
- The roots α and γ are from the equation \( A x^2 - 4x + 1 = 0 \).
- The roots β and δ are from the equation \( B x^2 - 6x + 1 = 0 \).
- The roots α, β, γ, and δ are in H.P.

Since α, β, γ, and δ are in H.P., the reciprocals \( \frac{1}{\alpha}, \frac{1}{\beta}, \frac{1}{\gamma}, \frac{1}{\delta} \) are in A.P.

### Step 2: Use Vieta's formulas
From Vieta's formulas:
- For the equation \( A x^2 - 4x + 1 = 0 \):
- Sum of roots: \( α + γ = \frac{4}{A} \)
- Product of roots: \( αγ = \frac{1}{A} \)

- For the equation \( B x^2 - 6x + 1 = 0 \):
- Sum of roots: \( β + δ = \frac{6}{B} \)
- Product of roots: \( βδ = \frac{1}{B} \)

### Step 3: Set up the equations for H.P.
Since \( \frac{1}{\alpha}, \frac{1}{\beta}, \frac{1}{\gamma}, \frac{1}{\delta} \) are in A.P., we can express them as:
- \( \frac{1}{\alpha} = a - 3d \)
- \( \frac{1}{\beta} = a - d \)
- \( \frac{1}{\gamma} = a + d \)
- \( \frac{1}{\delta} = a + 3d \)

### Step 4: Establish relationships using sums
From the H.P. condition:
1. The sum of the first and last terms equals the sum of the middle terms:
\[ (a - 3d) + (a + 3d) = (a - d) + (a + d) \]
Simplifying gives:
\[ 2a = 2a \quad \text{(This holds true)} \]

### Step 5: Find expressions for sums and products
Using the relationships we derived from Vieta’s formulas:
- For \( α + γ \):
\[ \frac{1}{\alpha} + \frac{1}{\gamma} = \frac{α + γ}{αγ} = \frac{4/A}{1/A} = 4 \implies (a - 3d) + (a + d) = 4 \]
Simplifying gives:
\[ 2a - 2d = 4 \quad \text{(Equation 1)} \]

- For \( β + δ \):
\[ \frac{1}{\beta} + \frac{1}{\delta} = \frac{β + δ}{βδ} = \frac{6/B}{1/B} = 6 \implies (a - d) + (a + 3d) = 6 \]
Simplifying gives:
\[ 2a + 2d = 6 \quad \text{(Equation 2)} \]

### Step 6: Solve the system of equations
Now we have two equations:
1. \( 2a - 2d = 4 \) (Equation 1)
2. \( 2a + 2d = 6 \) (Equation 2)

Adding these two equations:
\[ (2a - 2d) + (2a + 2d) = 4 + 6 \implies 4a = 10 \implies a = \frac{5}{2} \]

Substituting \( a \) back into Equation 1:
\[ 2\left(\frac{5}{2}\right) - 2d = 4 \implies 5 - 2d = 4 \implies 2d = 1 \implies d = \frac{1}{2} \]

### Step 7: Find values of A and B
Now we can find \( A \) and \( B \):
- For \( A \):
\[ A = \frac{1}{αγ} = \frac{1}{\left(\frac{1}{a - 3d}\right)\left(\frac{1}{a + d}\right)} = \frac{1}{\left(\frac{1}{\frac{5}{2} - \frac{3}{2}}\right)\left(\frac{1}{\frac{5}{2} + \frac{1}{2}}\right)} = \frac{1}{\left(\frac{1}{1}\right)\left(\frac{1}{3}\right)} = 3 \]

- For \( B \):
\[ B = \frac{1}{βδ} = \frac{1}{\left(\frac{1}{a - d}\right)\left(\frac{1}{a + 3d}\right)} = \frac{1}{\left(\frac{1}{\frac{5}{2} - \frac{1}{2}}\right)\left(\frac{1}{\frac{5}{2} + \frac{3}{2}}\right)} = \frac{1}{\left(\frac{1}{2}\right)\left(\frac{1}{4}\right)} = 8 \]

### Conclusion
Thus, the values are:
- \( A = 3 \)
- \( B = 8 \)

### Final Answer
The correct options are:
- \( A = 3 \) (Option a)
- \( B = 8 \) (Option d)