If `a_(1),a_(2),a_(3)…` are in G.P. then the value of
`|{:(log a_(n),loga_(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)):}|` is

Options

A. `log a_(1)a_(3)-2log a_(2)`
B. `-1`
C. 0
D. `log(a_(n)a_(n+8))-2loga_(n+4)`

Answer

To solve the problem, we need to find the value of 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:

\[ a_n = ar^{n-1}, \quad a_{n+1} = ar^n, \quad a_{n+2} = ar^{n+1}, \quad a_{n+3} = ar^{n+2}, \quad a_{n+4} = ar^{n+3}, \quad a_{n+5} = ar^{n+4}, \quad a_{n+6} = ar^{n+5}, \quad a_{n+7} = ar^{n+6}, \quad a_{n+8} = ar^{n+7} \]

Taking logarithms, we have:

\[ \log a_n = \log a + (n-1) \log r, \quad \log a_{n+1} = \log a + n \log r, \quad \log a_{n+2} = \log a + (n+1) \log r \]

Continuing this way, we can express all logarithmic terms in the determinant:

\[ \log a_{n+3} = \log a + (n+2) \log r, \quad \log a_{n+4} = \log a + (n+3) \log r, \quad \log a_{n+5} = \log a + (n+4) \log r \]

\[ \log a_{n+6} = \log a + (n+5) \log r, \quad \log a_{n+7} = \log a + (n+6) \log r, \quad \log a_{n+8} = \log a + (n+7) \log r \]

Substituting these into the determinant, we get:

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

Now, we can simplify the determinant by performing column operations. We will subtract the first column from the second and third columns:

1. **Column 2**: \(C_2 \to C_2 - C_1\)
2. **Column 3**: \(C_3 \to C_3 - C_1\)

This gives us:

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

This simplifies to:

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

Now, notice that the second and third columns are proportional, which means that the determinant is equal to zero.

Thus, the value of the determinant is:

\[ \boxed{0} \]