Suppose `alpha, gamma` are roots of the equation `ax^(2)-4x+1=0 and beta, delta` be the roots of the equation `bx^(2)-6x+1=0." If "(1)/(alpha),(1)/(beta),(1)/(gamma),(1)/(delta)` are in A.P., then a+b =

Answer

To solve the problem step by step, we need to find the values of \( a \) and \( b \) from the given quadratic equations and then calculate \( a + b \).

### Step 1: Identify the roots of the equations
We have two quadratic equations:
1. \( ax^2 - 4x + 1 = 0 \) with roots \( \alpha \) and \( \gamma \).
2. \( bx^2 - 6x + 1 = 0 \) with roots \( \beta \) and \( \delta \).

### Step 2: Use Vieta's formulas
From Vieta's formulas:
- For the first equation, the sum of the roots \( \alpha + \gamma = \frac{4}{a} \) and the product of the roots \( \alpha \cdot \gamma = \frac{1}{a} \).
- For the second equation, the sum of the roots \( \beta + \delta = \frac{6}{b} \) and the product of the roots \( \beta \cdot \delta = \frac{1}{b} \).

### Step 3: Set up the condition for the roots in A.P.
The condition states that \( \frac{1}{\alpha}, \frac{1}{\beta}, \frac{1}{\gamma}, \frac{1}{\delta} \) are in arithmetic progression (A.P.). For four numbers \( A, B, C, D \) to be in A.P., the condition is:
\[ 2B = A + C \]

Here, let:
- \( A = \frac{1}{\alpha} \)
- \( B = \frac{1}{\beta} \)
- \( C = \frac{1}{\gamma} \)
- \( D = \frac{1}{\delta} \)

Thus, we have:
\[ 2 \cdot \frac{1}{\beta} = \frac{1}{\alpha} + \frac{1}{\gamma} \]

### Step 4: Substitute the values from Vieta's formulas
Substituting the values:
\[ 2 \cdot \frac{1}{\beta} = \frac{1}{\alpha} + \frac{1}{\gamma} \]
\[ 2 \cdot \frac{1}{\beta} = \frac{\alpha + \gamma}{\alpha \cdot \gamma} \]
Using Vieta's results:
\[ 2 \cdot \frac{1}{\beta} = \frac{\frac{4}{a}}{\frac{1}{a}} = 4 \]
Thus, we have:
\[ \frac{2}{\beta} = 4 \]
This gives:
\[ \beta = \frac{1}{2} \]

### Step 5: Find \( \delta \) using the product of the roots
Using the product of the roots:
\[ \beta \cdot \delta = \frac{1}{b} \]
Substituting \( \beta = \frac{1}{2} \):
\[ \frac{1}{2} \cdot \delta = \frac{1}{b} \]
This implies:
\[ \delta = \frac{2}{b} \]

### Step 6: Use the sum of the roots to find \( b \)
Using the sum of the roots:
\[ \beta + \delta = \frac{6}{b} \]
Substituting \( \beta = \frac{1}{2} \) and \( \delta = \frac{2}{b} \):
\[ \frac{1}{2} + \frac{2}{b} = \frac{6}{b} \]
Multiplying through by \( 2b \) to eliminate the fractions:
\[ b + 4 = 12 \]
Thus:
\[ b = 8 \]

### Step 7: Find \( a \) using the product of the roots for \( \alpha \) and \( \gamma \)
From the product of the roots:
\[ \alpha \cdot \gamma = \frac{1}{a} \]
Using \( \alpha \cdot \gamma = 1 \) (from the earlier calculations):
\[ 1 = \frac{1}{a} \]
Thus:
\[ a = 3 \]

### Step 8: Calculate \( a + b \)
Now we have:
- \( a = 3 \)
- \( b = 8 \)

So:
\[ a + b = 3 + 8 = 11 \]

### Final Answer
Thus, the final answer is:
\[ \boxed{11} \]