If `a_1,a_2,a_3,….,a_n` are in G.P. are in `a_i > 0` for each I, then the determinant
`Delta=|{:(loga_n,log a_(n+2),log a_(n+4)),(log a_(n+6),log a_(n+8), log a_(n+10)),(log a_(n+12), log a_(n+14), loga_(n+16)):}|`

Options

A. 0
B. `(sum_(i=1)^(n^2+n) a_i)`
C. 1
D. 2

Answer

To solve the determinant given in the problem, we start with the determinant:

\[ \Delta = \begin{vmatrix} \log a_n & \log a_{n+2} & \log a_{n+4} \\ \log a_{n+6} & \log a_{n+8} & \log a_{n+10} \\ \log a_{n+12} & \log a_{n+14} & \log a_{n+16} \end{vmatrix} \]

### Step 1: Use properties of logarithms
Since \(a_1, a_2, a_3, \ldots, a_n\) are in geometric progression (G.P.), we can express \(a_k\) in terms of \(a_1\) and the common ratio \(r\):
\[ a_k = a_1 r^{k-1} \]

Thus, we can rewrite the logarithms:
\[ \log a_k = \log a_1 + (k-1) \log r \]

### Step 2: Substitute into the determinant
Substituting this into the determinant gives:
\[ \Delta = \begin{vmatrix} \log a_1 + (n-1) \log r & \log a_1 + (n+1) \log r & \log a_1 + (n+3) \log r \\ \log a_1 + (n+5) \log r & \log a_1 + (n+7) \log r & \log a_1 + (n+9) \log r \\ \log a_1 + (n+11) \log r & \log a_1 + (n+13) \log r & \log a_1 + (n+15) \log r \end{vmatrix} \]

### Step 3: Factor out common terms
We can factor out \(\log a_1\) and \(\log r\) from each column:
\[ \Delta = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{vmatrix} \cdot \log r^3 \]

### Step 4: Simplify the determinant
The determinant of a matrix where all rows (or columns) are identical is zero:
\[ \Delta = 0 \]

### Conclusion
Thus, the value of the determinant \(\Delta\) is:
\[ \Delta = 0 \]