If `a_1,a_2,…..a_n` are in G.P. then
`Delta = |("log"a_(n),"log" a_(n + 1),"log" a_(n + 2)),("log"a_(n + 3),"log" a_(n + 4),"log" a_(n + 5)),("log"a_(n + 6),"log" a_(n + 7),"log"a_(n + 8))|`

Options

A. 4
B. 2
C. 1
D. 0

Answer

To solve the problem, we need to evaluate the determinant given that \( a_1, a_2, \ldots, a_n \) are in a geometric progression (G.P.).

### Step-by-Step Solution:

1. **Understanding the terms in G.P.**:
Since \( a_1, a_2, \ldots, a_n \) are in G.P., we can express the terms as:
\[ a_n = a \cdot r^{n-1}, \quad a_{n+1} = a \cdot r^n, \quad a_{n+2} = a \cdot r^{n+1}, \quad a_{n+3} = a \cdot r^{n+2}, \quad a_{n+4} = a \cdot r^{n+3}, \quad a_{n+5} = a \cdot r^{n+4}, \quad a_{n+6} = a \cdot r^{n+5}, \quad a_{n+7} = a \cdot r^{n+6}, \quad a_{n+8} = a \cdot r^{n+7} \]

2. **Setting up the determinant**:
The determinant we need to evaluate is:
\[ \Delta = \begin{vmatrix} \log a_n & \log a_{n+1} & \log a_{n+2} \\ \log a_{n+3} & \log a_{n+4} & \log a_{n+5} \\ \log a_{n+6} & \log a_{n+7} & \log a_{n+8} \end{vmatrix} \]

Substituting the expressions for \( a_n, a_{n+1}, \ldots, a_{n+8} \):
\[ \Delta = \begin{vmatrix} \log(a \cdot r^{n-1}) & \log(a \cdot r^n) & \log(a \cdot r^{n+1}) \\ \log(a \cdot r^{n+2}) & \log(a \cdot r^{n+3}) & \log(a \cdot r^{n+4}) \\ \log(a \cdot r^{n+5}) & \log(a \cdot r^{n+6}) & \log(a \cdot r^{n+7}) \end{vmatrix} \]

3. **Using logarithmic properties**:
Using the property \( \log(ab) = \log a + \log b \):
\[ \Delta = \begin{vmatrix} \log a + (n-1) \log r & \log a + n \log r & \log a + (n+1) \log r \\ \log a + (n+2) \log r & \log a + (n+3) \log r & \log a + (n+4) \log r \\ \log a + (n+5) \log r & \log a + (n+6) \log r & \log a + (n+7) \log r \end{vmatrix} \]

4. **Simplifying the determinant**:
Let \( \log a = A \) and \( \log r = B \):
\[ \Delta = \begin{vmatrix} A + (n-1)B & A + nB & A + (n+1)B \\ A + (n+2)B & A + (n+3)B & A + (n+4)B \\ A + (n+5)B & A + (n+6)B & A + (n+7)B \end{vmatrix} \]

This can be rewritten as:
\[ \Delta = \begin{vmatrix} A & A & A \\ A & A & A \\ A & A & A \end{vmatrix} + B \cdot \begin{vmatrix} n-1 & n & n+1 \\ n+2 & n+3 & n+4 \\ n+5 & n+6 & n+7 \end{vmatrix} \]

5. **Identifying linear dependence**:
The first determinant is zero because all rows are identical. Therefore:
\[ \Delta = B \cdot \begin{vmatrix} n-1 & n & n+1 \\ n+2 & n+3 & n+4 \\ n+5 & n+6 & n+7 \end{vmatrix} \]

The second determinant can be shown to be zero as well because the rows are linearly dependent (the differences between the rows are constant).

6. **Conclusion**:
Since both determinants evaluate to zero, we conclude that:
\[ \Delta = 0 \]

### Final Answer:
\[ \Delta = 0 \]