If `x_(1),x_(2),x_(3),……………..x_(13)` are in A.P. then the value of
`|(e^(x_(1)),e^(x_(4)),e^(x^(7))),(e^(x_(4)),e^(x_(7)),e^(x_(10))),(e^(x_(7)),e^(x_(10)),e^(x_(13)))|` is

Options

A. 27
B. 0
C. 1
D. 9

Answer

To solve the given problem, we need to evaluate the determinant:

\[ D = \begin{vmatrix} e^{x_1} & e^{x_4} & e^{x_7} \\ e^{x_4} & e^{x_7} & e^{x_{10}} \\ e^{x_7} & e^{x_{10}} & e^{x_{13}} \end{vmatrix} \]

Given that \(x_1, x_2, x_3, \ldots, x_{13}\) are in Arithmetic Progression (A.P.), we can express these terms as follows:

- Let \(x_1 = a\) (the first term)
- Then, \(x_2 = a + d\), \(x_3 = a + 2d\), ..., \(x_{13} = a + 12d\)

Now we can substitute the values of \(x_1, x_4, x_7, x_{10}, x_{13}\) into the determinant:

- \(x_1 = a\)
- \(x_4 = a + 3d\)
- \(x_7 = a + 6d\)
- \(x_{10} = a + 9d\)
- \(x_{13} = a + 12d\)

Thus, we can rewrite the determinant as:

\[ D = \begin{vmatrix} e^a & e^{a + 3d} & e^{a + 6d} \\ e^{a + 3d} & e^{a + 6d} & e^{a + 9d} \\ e^{a + 6d} & e^{a + 9d} & e^{a + 12d} \end{vmatrix} \]

Next, we can factor out \(e^a\), \(e^{a + 3d}\), and \(e^{a + 6d}\) from each row:

\[ D = e^{a} \cdot e^{a + 3d} \cdot e^{a + 6d} \begin{vmatrix} 1 & e^{3d} & e^{6d} \\ e^{3d} & e^{6d} & e^{9d} \\ e^{6d} & e^{9d} & e^{12d} \end{vmatrix} \]

Now, we can simplify the determinant further:

\[ D = e^{3a + 9d} \begin{vmatrix} 1 & e^{3d} & e^{6d} \\ e^{3d} & e^{6d} & e^{9d} \\ e^{6d} & e^{9d} & e^{12d} \end{vmatrix} \]

Next, we observe that each row of the determinant is proportional to the others. Specifically, if we multiply the first row by \(e^{3d}\), we get the second row, and if we multiply the second row by \(e^{3d}\), we get the third row. This means that the rows are linearly dependent.

Thus, the value of the determinant is:

\[ D = 0 \]

**Final Answer:**
The value of the determinant is \(0\).

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