If `T_p` be the pth term of a G.P. of all +ive terms then
`Delta = |("log"T_(P + 1),"log" T_(P + 3),"log" T_(P + 5)),("log"T_(P + 3),"log" T_(P + 5),"log" T_(P + 7)),("log"T_(P + 5),"log" T_(P + 7),"log" T_(P + 9))|`
is equal to

Options

A. 0
B. 1
C. 2
D. none of these

Answer

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

\[ \Delta = \begin{vmatrix} \log T_{P + 1} & \log T_{P + 3} & \log T_{P + 5} \\ \log T_{P + 3} & \log T_{P + 5} & \log T_{P + 7} \\ \log T_{P + 5} & \log T_{P + 7} & \log T_{P + 9} \end{vmatrix} \]

### Step 1: Express the terms of the G.P.
Let the \( p \)-th term of the G.P. be represented as:
\[ T_p = ar^{p-1} \]
where \( a \) is the first term and \( r \) is the common ratio. Therefore, we can express the logarithmic terms as follows:
\[ \log T_{P + 1} = \log(ar^P) = \log a + P \log r \]
\[ \log T_{P + 3} = \log(ar^{P + 2}) = \log a + (P + 2) \log r \]
\[ \log T_{P + 5} = \log(ar^{P + 4}) = \log a + (P + 4) \log r \]
\[ \log T_{P + 7} = \log(ar^{P + 6}) = \log a + (P + 6) \log r \]
\[ \log T_{P + 9} = \log(ar^{P + 8}) = \log a + (P + 8) \log r \]

### Step 2: Substitute the logarithmic terms into the determinant
Now substituting these expressions into the determinant, we have:
\[ \Delta = \begin{vmatrix} \log a + P \log r & \log a + (P + 2) \log r & \log a + (P + 4) \log r \\ \log a + (P + 2) \log r & \log a + (P + 4) \log r & \log a + (P + 6) \log r \\ \log a + (P + 4) \log r & \log a + (P + 6) \log r & \log a + (P + 8) \log r \end{vmatrix} \]

### Step 3: Simplify the determinant
Notice that we can factor out \(\log r\) from each column:
\[ \Delta = \log r \begin{vmatrix} 1 & P & P + 2 \\ 1 & P + 2 & P + 4 \\ 1 & P + 4 & P + 6 \end{vmatrix} \]

### Step 4: Calculate the determinant
Now we can simplify the determinant:
\[ \Delta = \log r \begin{vmatrix} 1 & P & P + 2 \\ 1 & P + 2 & P + 4 \\ 1 & P + 4 & P + 6 \end{vmatrix} \]

Perform row operations (subtract the first row from the second and the third):
\[ = \begin{vmatrix} 1 & P & P + 2 \\ 0 & 2 & 2 \\ 0 & 2 & 2 \end{vmatrix} \]

Now, we see that the second and third rows are identical, which means the determinant is zero:
\[ \Delta = 0 \]

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