If `a_(1), a_(2), a_(3), a_(4), a_(5)` are consecutive terms of an arithmetic progression with common difference 3, then the value of `|(a_(3)^(2),a_(2),a_(1)),(a_(4)^(2),a_(3),a_(2)),(a_(5)^(2),a_(4),a_(3))|` is

Options

A. 0
B. 27
C. 81
D. 162

Answer

To solve the problem, we need to evaluate the determinant given the terms of an arithmetic progression (AP). Let's break down the steps:

### Step 1: Define the terms of the AP
Let the first term of the arithmetic progression be \( a_1 \). Since the common difference is 3, we can express the terms as follows:
- \( a_1 = a_1 \)
- \( a_2 = a_1 + 3 \)
- \( a_3 = a_1 + 6 \)
- \( a_4 = a_1 + 9 \)
- \( a_5 = a_1 + 12 \)

### Step 2: Write the determinant
We need to evaluate the determinant:
\[ D = \begin{vmatrix} a_3^2 & a_2 & a_1 \\ a_4^2 & a_3 & a_2 \\ a_5^2 & a_4 & a_3 \end{vmatrix} \]

### Step 3: Substitute the terms into the determinant
Substituting the expressions for \( a_1, a_2, a_3, a_4, a_5 \):
- \( a_3^2 = (a_1 + 6)^2 = a_1^2 + 12a_1 + 36 \)
- \( a_4^2 = (a_1 + 9)^2 = a_1^2 + 18a_1 + 81 \)
- \( a_5^2 = (a_1 + 12)^2 = a_1^2 + 24a_1 + 144 \)

Thus, the determinant becomes:
\[ D = \begin{vmatrix} a_1^2 + 12a_1 + 36 & a_1 + 3 & a_1 \\ a_1^2 + 18a_1 + 81 & a_1 + 6 & a_1 + 3 \\ a_1^2 + 24a_1 + 144 & a_1 + 9 & a_1 + 6 \end{vmatrix} \]

### Step 4: Simplify the determinant using row operations
We can use properties of determinants to simplify. We will perform the operation \( C_2 \leftarrow C_2 - C_3 \):
\[ D = \begin{vmatrix} a_1^2 + 12a_1 + 36 & 3 & a_1 \\ a_1^2 + 18a_1 + 81 & 3 & a_1 + 3 \\ a_1^2 + 24a_1 + 144 & 3 & a_1 + 6 \end{vmatrix} \]

### Step 5: Factor out common terms
Now, we can factor out the common term 3 from the second column:
\[ D = 3 \begin{vmatrix} a_1^2 + 12a_1 + 36 & 1 & a_1 \\ a_1^2 + 18a_1 + 81 & 1 & a_1 + 3 \\ a_1^2 + 24a_1 + 144 & 1 & a_1 + 6 \end{vmatrix} \]

### Step 6: Use row operations again
Next, we can perform \( R_2 \leftarrow R_2 - R_1 \) and \( R_3 \leftarrow R_3 - R_1 \):
\[ D = 3 \begin{vmatrix} a_1^2 + 12a_1 + 36 & 1 & a_1 \\ 6a_1 + 45 & 0 & 3 \\ 12a_1 + 108 & 0 & 6 \end{vmatrix} \]

### Step 7: Calculate the determinant
Now, we can calculate the determinant:
\[ D = 3 \cdot 3 \cdot \begin{vmatrix} 1 & a_1 \\ 0 & 6 \end{vmatrix} = 3 \cdot 3 \cdot (1 \cdot 6 - 0 \cdot a_1) = 9 \cdot 6 = 54 \]

### Step 8: Substitute back the common difference
Since the common difference \( D = 3 \), we can substitute back:
\[ D = 54 \cdot D^2 = 54 \cdot 3^2 = 54 \cdot 9 = 486 \]

### Final Answer
The value of the determinant is \( 486 \).