If `a_1, a_2, a_3,.......` are in G.P. then the value of determinant `|(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)))|` equals (A) 0 (B) 1 (C) 2 (D) 3

Options

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

Explanation

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

\[ D = \begin{vmatrix} \log(a_n) & \log(a_{n+1}) & \log(a_{n+2}) \\ \log(a_{n+3}) & \log(a_{n+4}) & \log(a_{n+5}) \\ \log(a_{n+6}) & \log(a_{n+7}) & \log(a_{n+8}) \end{vmatrix} \]

Given that \(a_1, a_2, a_3, \ldots\) are in a geometric progression (G.P.), we can express the terms as follows:

\[ a_n = a \cdot r^{n-1}, \quad a_{n+k} = a \cdot r^{n+k-1} \quad \text{for } k = 0, 1, 2, \ldots \]

where \(a\) is the first term and \(r\) is the common ratio.

### Step 1: Rewrite the logarithmic terms
Using the properties of logarithms, we can rewrite the logarithmic terms:

\[ \log(a_n) = \log(a) + (n-1) \log(r) \]
\[ \log(a_{n+k}) = \log(a) + (n+k-1) \log(r) \quad \text{for } k = 0, 1, 2, \ldots, 8 \]

### Step 2: Substitute into the determinant
Substituting these into the determinant gives:

\[ D = \begin{vmatrix} \log(a) + (n-1) \log(r) & \log(a) + n \log(r) & \log(a) + (n+1) \log(r) \\ \log(a) + (n+2) \log(r) & \log(a) + (n+3) \log(r) & \log(a) + (n+4) \log(r) \\ \log(a) + (n+5) \log(r) & \log(a) + (n+6) \log(r) & \log(a) + (n+7) \log(r) \end{vmatrix} \]

### Step 3: Factor out common terms
We can factor out the common terms from each column:

\[ D = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{vmatrix} \cdot \log(a) + \begin{vmatrix} (n-1) & n & n+1 \\ (n+2) & (n+3) & (n+4) \\ (n+5) & (n+6) & (n+7) \end{vmatrix} \cdot \log(r) \]

### Step 4: Analyze the determinant
Notice that the first part of the determinant simplifies to zero because all rows are identical. Therefore, we only need to evaluate the second determinant.

### Step 5: Evaluate the second determinant
The second determinant can be simplified as follows:

\[ D' = \begin{vmatrix} (n-1) & n & n+1 \\ (n+2) & (n+3) & (n+4) \\ (n+5) & (n+6) & (n+7) \end{vmatrix} \]

Using properties of determinants, we can see that the rows are also in arithmetic progression, which means that the determinant will also evaluate to zero.

### Conclusion
Thus, the value of the original determinant \(D\) is:

\[ D = 0 \]

Therefore, the answer is (A) 0.