If `2^(a1),2^(a2),2^(a3).......2^(ar)` are in geometric progression , then `|{:(a_1,a_2,a_3),(a_(n+1),a_(n+2),a_(n+3)),(a_(2n+1),a_(2n+2),a_(2n+3)):}|` is equal to

Options

A. `2^5`
B. `2^3`
C. 0
D. None of these

Answer

To solve the problem, we start by analyzing the given geometric progression of the terms \(2^{a_1}, 2^{a_2}, 2^{a_3}, \ldots, 2^{a_r}\).

### Step 1: Understand the condition for a geometric progression
For the terms \(2^{a_1}, 2^{a_2}, 2^{a_3}, \ldots, 2^{a_r}\) to be in geometric progression, the condition that must hold is:
\[ 2^{a_2}^2 = 2^{a_1} \cdot 2^{a_3} \]
This simplifies to:
\[ 2^{2a_2} = 2^{a_1 + a_3} \]
Taking logarithm base 2 on both sides gives:
\[ 2a_2 = a_1 + a_3 \]
This implies that \(a_1, a_2, a_3\) are in arithmetic progression (AP).

### Step 2: Generalize for all terms
Since the same condition holds for all consecutive terms in the sequence, we can conclude that:
\[ a_k = a_1 + (k-1)d \]
for some common difference \(d\). Hence, \(a_1, a_2, \ldots, a_r\) are in AP.

### Step 3: Set up the determinant
We need to evaluate the determinant:
\[ D = \begin{vmatrix} a_1 & a_2 & a_3 \\ a_{n+1} & a_{n+2} & a_{n+3} \\ a_{2n+1} & a_{2n+2} & a_{2n+3} \end{vmatrix} \]

### Step 4: Express the terms in terms of \(a_1\) and \(d\)
Using the AP property:
- \(a_{n+1} = a_1 + nd\)
- \(a_{n+2} = a_1 + (n+1)d\)
- \(a_{n+3} = a_1 + (n+2)d\)
- \(a_{2n+1} = a_1 + 2nd\)
- \(a_{2n+2} = a_1 + (2n+1)d\)
- \(a_{2n+3} = a_1 + (2n+2)d\)

### Step 5: Substitute into the determinant
Substituting these values into the determinant gives:
\[ D = \begin{vmatrix} a_1 & a_1 + d & a_1 + 2d \\ a_1 + nd & a_1 + (n+1)d & a_1 + (n+2)d \\ a_1 + 2nd & a_1 + (2n+1)d & a_1 + (2n+2)d \end{vmatrix} \]

### Step 6: Row operations
Now, we can perform row operations. If we subtract the first row from the second and third rows, we get:
\[ D = \begin{vmatrix} a_1 & a_1 + d & a_1 + 2d \\ nd & d & 2d \\ 2nd & nd & (2n+1)d \end{vmatrix} \]

### Step 7: Factor out common terms
We can factor out \(d\) from the second and third rows:
\[ D = d^2 \begin{vmatrix} a_1 & a_1 + d & a_1 + 2d \\ n & 1 & 2 \\ 2n & n & (2n+1) \end{vmatrix} \]

### Step 8: Evaluate the determinant
Notice that if we perform another row operation (subtracting \(2\) times the second row from the third), we will find that the second and third rows become linearly dependent, which means:
\[ D = 0 \]

### Conclusion
Thus, the value of the determinant is:
\[ \boxed{0} \]