Let a, b be the roots of the equation ` x^(2) - 4 x +k_(1) = 0`
and c , d the roots of the equation
` x^(2) - 36 x + k_(2) = 0 ` If ` a lt b lt c lt d and a, b,c,d ` are in G.P. then the product ` k_(1) k_(2)` equals

Options

A. 81
B. 729
C. 265
D. 64

Answer

To solve the problem, we need to find the product \( k_1 k_2 \) given the roots of two quadratic equations and the condition that the roots are in geometric progression (G.P.).

### Step-by-Step Solution:

1. **Identify the Roots and Their Relationships**:
- Let \( a \) and \( b \) be the roots of the equation \( x^2 - 4x + k_1 = 0 \).
- Let \( c \) and \( d \) be the roots of the equation \( x^2 - 36x + k_2 = 0 \).
- Given that \( a < b < c < d \) and \( a, b, c, d \) are in G.P., we can express the roots as:
- \( a = \alpha \)
- \( b = \alpha r \)
- \( c = \alpha r^2 \)
- \( d = \alpha r^3 \)

2. **Use Vieta's Formulas**:
- From the first equation \( x^2 - 4x + k_1 = 0 \):
- The sum of the roots \( a + b = 4 \) gives:
\[ \alpha + \alpha r = 4 \implies \alpha(1 + r) = 4 \quad \text{(Equation 1)} \]
- The product of the roots \( ab = k_1 \) gives:
\[ \alpha \cdot \alpha r = k_1 \implies \alpha^2 r = k_1 \quad \text{(Equation 2)} \]

- From the second equation \( x^2 - 36x + k_2 = 0 \):
- The sum of the roots \( c + d = 36 \) gives:
\[ \alpha r^2 + \alpha r^3 = 36 \implies \alpha r^2(1 + r) = 36 \quad \text{(Equation 3)} \]
- The product of the roots \( cd = k_2 \) gives:
\[ \alpha r^2 \cdot \alpha r^3 = k_2 \implies \alpha^2 r^5 = k_2 \quad \text{(Equation 4)} \]

3. **Relate Equations**:
- From Equation 1, we have:
\[ \alpha(1 + r) = 4 \implies \alpha = \frac{4}{1 + r} \]
- Substitute \( \alpha \) into Equation 3:
\[ \frac{4}{1 + r} r^2(1 + r) = 36 \implies 4r^2 = 36 \implies r^2 = 9 \implies r = 3 \quad (\text{since } r > 0) \]

4. **Find \( \alpha \)**:
- Substitute \( r = 3 \) back into Equation 1:
\[ \alpha(1 + 3) = 4 \implies 4\alpha = 4 \implies \alpha = 1 \]

5. **Calculate \( k_1 \) and \( k_2 \)**:
- From Equation 2:
\[ k_1 = \alpha^2 r = 1^2 \cdot 3 = 3 \]
- From Equation 4:
\[ k_2 = \alpha^2 r^5 = 1^2 \cdot 3^5 = 243 \]

6. **Find the Product \( k_1 k_2 \)**:
- Calculate \( k_1 k_2 \):
\[ k_1 k_2 = 3 \cdot 243 = 729 \]

### Final Answer:
Thus, the product \( k_1 k_2 \) is \( \boxed{729} \).