If ` alpha , beta ` are the roots of ` x^2 - 3x +a =0` and ` gamma , delta ` are the roots of ` x^2 -12 x+b=0` and ` alpha , beta , gamma , delta ` in that order from a geometric progression increasing order with common ration ` r gt 1` then ` a+b=`

Options

A. 16
B. 28
C. 34
D. 42

Answer

To solve the problem, we need to find the values of \( a \) and \( b \) given the equations and the conditions about the roots being in a geometric progression. Let's go through the solution step by step.

### Step 1: Identify the roots
Let the roots of the first equation \( x^2 - 3x + a = 0 \) be \( \alpha \) and \( \beta \). According to Vieta's formulas:
- The sum of the roots \( \alpha + \beta = 3 \)
- The product of the roots \( \alpha \beta = a \)

Let the roots of the second equation \( x^2 - 12x + b = 0 \) be \( \gamma \) and \( \delta \). Again, by Vieta's formulas:
- The sum of the roots \( \gamma + \delta = 12 \)
- The product of the roots \( \gamma \delta = b \)

### Step 2: Express the roots in terms of a common ratio
Since \( \alpha, \beta, \gamma, \delta \) are in a geometric progression with a common ratio \( r > 1 \), we can express them as:
- \( \alpha = a \)
- \( \beta = ar \)
- \( \gamma = ar^2 \)
- \( \delta = ar^3 \)

### Step 3: Set up equations based on the sums of the roots
From the sum of the roots:
1. For \( \alpha + \beta = 3 \):
\[ a + ar = 3 \implies a(1 + r) = 3 \quad \text{(Equation 1)} \]

2. For \( \gamma + \delta = 12 \):
\[ ar^2 + ar^3 = 12 \implies ar^2(1 + r) = 12 \quad \text{(Equation 2)} \]

### Step 4: Divide Equation 2 by Equation 1
Dividing Equation 2 by Equation 1 gives:
\[ \frac{ar^2(1 + r)}{a(1 + r)} = \frac{12}{3} \]
This simplifies to:
\[ r^2 = 4 \implies r = 2 \quad \text{(since \( r > 1 \))} \]

### Step 5: Substitute \( r \) back to find \( a \)
Substituting \( r = 2 \) into Equation 1:
\[ a(1 + 2) = 3 \implies 3a = 3 \implies a = 1 \]

### Step 6: Find \( b \)
Now, we substitute \( a = 1 \) and \( r = 2 \) into the expressions for \( \gamma \) and \( \delta \):
- \( \gamma = ar^2 = 1 \cdot 2^2 = 4 \)
- \( \delta = ar^3 = 1 \cdot 2^3 = 8 \)

Now we can find \( b \):
\[ \gamma \delta = b \implies 4 \cdot 8 = b \implies b = 32 \]

### Step 7: Calculate \( a + b \)
Finally, we find \( a + b \):
\[ a + b = 1 + 32 = 33 \]

Thus, the final answer is:
\[ \boxed{33} \]