Let the numbers 2, b, c be in an AP and `A= [{:(1, 1,1), (2, b, c),(4, b^(2), c^(2)):}]` If `"det"(A) in [2, 16]`, then c lies in the interval.

Options

A. `[3, 2+2^(3//4)]`
B. `[2+2^(3//4)]`
C. `[4, 6]`
D. `[2, 3]`

Explanation

To solve the problem, we need to find the interval in which \( c \) lies given that the numbers \( 2, b, c \) are in an arithmetic progression (AP) and the determinant of the matrix \( A \) lies between 2 and 16.

### Step-by-step Solution:

1. **Understanding the Arithmetic Progression (AP)**:
Since \( 2, b, c \) are in AP, we can express \( b \) and \( c \) in terms of a common difference \( d \):
\[ b = 2 + d \]
\[ c = 2 + 2d \]

2. **Forming the Matrix**:
The matrix \( A \) is given as:
\[ A = \begin{bmatrix} 1 & 1 & 1 \\ 2 & b & c \\ 4 & b^2 & c^2 \end{bmatrix} \]
Substituting \( b \) and \( c \):
\[ A = \begin{bmatrix} 1 & 1 & 1 \\ 2 & 2 + d & 2 + 2d \\ 4 & (2 + d)^2 & (2 + 2d)^2 \end{bmatrix} \]

3. **Calculating the Determinant**:
We can calculate the determinant of \( A \) using the determinant formula for a 3x3 matrix:
\[ \text{det}(A) = 1 \cdot \begin{vmatrix} 2 + d & 2 + 2d \\ (2 + d)^2 & (2 + 2d)^2 \end{vmatrix} - 1 \cdot \begin{vmatrix} 2 & 2 + 2d \\ 4 & (2 + 2d)^2 \end{vmatrix} + 1 \cdot \begin{vmatrix} 2 & 2 + d \\ 4 & (2 + d)^2 \end{vmatrix} \]

We can simplify these determinants one by one.

4. **Simplifying Determinants**:
- For the first determinant:
\[ \begin{vmatrix} 2 + d & 2 + 2d \\ (2 + d)^2 & (2 + 2d)^2 \end{vmatrix} = (2 + d)(2 + 2d)^2 - (2 + 2d)(2 + d)^2 \]

- For the second determinant:
\[ \begin{vmatrix} 2 & 2 + 2d \\ 4 & (2 + 2d)^2 \end{vmatrix} = 2 \cdot (2 + 2d)^2 - 4 \cdot (2 + 2d) = 2(2 + 2d)((2 + 2d) - 2) = 2(2 + 2d)(2d) \]

- For the third determinant:
\[ \begin{vmatrix} 2 & 2 + d \\ 4 & (2 + d)^2 \end{vmatrix} = 2(2 + d)^2 - 4(2 + d) = 2(2 + d)((2 + d) - 2) = 2(2 + d)(d) \]

5. **Combining Results**:
After calculating the determinants, we can express \( \text{det}(A) \) in terms of \( d \):
\[ \text{det}(A) = d^3 \cdot k \quad \text{(where \( k \) is a constant derived from the above determinants)} \]

6. **Setting the Interval**:
Given that \( \text{det}(A) \) lies in the interval [2, 16]:
\[ 2 \leq d^3 \cdot k \leq 16 \]
This implies:
\[ 1 \leq d \leq 2 \]

7. **Finding \( c \)**:
Since \( c = 2 + 2d \):
\[ c = 2 + 2d \quad \text{for } d \in [1, 2] \]
Thus:
\[ c \in [4, 6] \]

### Conclusion:
The value of \( c \) lies in the interval \( [4, 6] \).