If `a_(1),a_(2),a_(3),……….a_(n)` ………… are in G.P and `a_(i) gt 0 ` then the value of the determinant
`|(loga_(n),log a_(n+1),loga_(n+2)),(loga_(n+1),loga_(n+2),loga_(n+3)),(loga_(n+2),loga_(n+3),loga_(n+4))|` is

Options

A. 2
B. 1
C. 0
D. `-2`

Answer

To solve the given determinant problem step by step, we start with the determinant:

\[ D = \begin{vmatrix} \log a_n & \log a_{n+1} & \log a_{n+2} \\ \log a_{n+1} & \log a_{n+2} & \log a_{n+3} \\ \log a_{n+2} & \log a_{n+3} & \log a_{n+4} \end{vmatrix} \]

### Step 1: Understanding the terms in the determinant
Since \(a_1, a_2, a_3, \ldots, a_n\) are in a geometric progression (G.P.), we can express the terms as:
- \(a_n = a_1 r^{n-1}\)
- \(a_{n+1} = a_1 r^n\)
- \(a_{n+2} = a_1 r^{n+1}\)
- \(a_{n+3} = a_1 r^{n+2}\)
- \(a_{n+4} = a_1 r^{n+3}\)

where \(r\) is the common ratio of the G.P.

### Step 2: Substitute the terms into the determinant
Now substituting these expressions into the determinant, we have:

\[ D = \begin{vmatrix} \log(a_1 r^{n-1}) & \log(a_1 r^n) & \log(a_1 r^{n+1}) \\ \log(a_1 r^n) & \log(a_1 r^{n+1}) & \log(a_1 r^{n+2}) \\ \log(a_1 r^{n+1}) & \log(a_1 r^{n+2}) & \log(a_1 r^{n+3}) \end{vmatrix} \]

Using the property of logarithms, we can simplify:

\[ \log(a_1 r^k) = \log a_1 + k \log r \]

Thus, we rewrite the determinant as:

\[ D = \begin{vmatrix} \log a_1 + (n-1) \log r & \log a_1 + n \log r & \log a_1 + (n+1) \log r \\ \log a_1 + n \log r & \log a_1 + (n+1) \log r & \log a_1 + (n+2) \log r \\ \log a_1 + (n+1) \log r & \log a_1 + (n+2) \log r & \log a_1 + (n+3) \log r \end{vmatrix} \]

### Step 3: Factor out common terms
We can factor out \(\log a_1\) and \(\log r\) from each column:

\[ D = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{vmatrix} \cdot (\log r)^2 \]

### Step 4: Analyze the determinant
Notice that the matrix formed by the coefficients of \(\log a_1\) and \(\log r\) is:

\[ \begin{vmatrix} n-1 & n & n+1 \\ n & n+1 & n+2 \\ n+1 & n+2 & n+3 \end{vmatrix} \]

This matrix has identical rows, which means that the determinant evaluates to zero.

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

\[ D = 0 \]