If `agt0 , b gt 0 , cgt0` are respectively the pth , qth , rth terms of G.P., then value of the determinant `|(loga, p ,1),(logb,q,1),(log c,r,1)|` is

Options

A. 0
B. 1
C. `-1`
D. none of these

Answer

To solve the problem, we need to evaluate the determinant

\[ D = \begin{vmatrix} \log a & p & 1 \\ \log b & q & 1 \\ \log c & r & 1 \end{vmatrix} \]

where \( a, b, c > 0 \) are the \( p \)-th, \( q \)-th, and \( r \)-th terms of a geometric progression (G.P.).

### Step 1: Express the terms of the G.P.
Let the first term of the G.P. be \( A \) and the common ratio be \( r \). The \( n \)-th term of a G.P. is given by:

\[ T_n = A \cdot r^{n-1} \]

Thus, we can express \( a, b, c \) as follows:

- \( a = A \cdot r^{p-1} \) (for the \( p \)-th term)
- \( b = A \cdot r^{q-1} \) (for the \( q \)-th term)
- \( c = A \cdot r^{r-1} \) (for the \( r \)-th term)

### Step 2: Substitute the values into the determinant
Now, substituting \( a, b, c \) into the determinant:

\[ D = \begin{vmatrix} \log(A \cdot r^{p-1}) & p & 1 \\ \log(A \cdot r^{q-1}) & q & 1 \\ \log(A \cdot r^{r-1}) & r & 1 \end{vmatrix} \]

Using the logarithmic property \( \log(xy) = \log x + \log y \), we can rewrite the determinant:

\[ D = \begin{vmatrix} \log A + (p-1) \log r & p & 1 \\ \log A + (q-1) \log r & q & 1 \\ \log A + (r-1) \log r & r & 1 \end{vmatrix} \]

### Step 3: Simplify the determinant
Now, we can perform row operations to simplify the determinant. We can subtract the first row from the second and third rows:

\[ D = \begin{vmatrix} \log A + (p-1) \log r & p & 1 \\ (q - p) \log r & q - p & 0 \\ (r - p) \log r & r - p & 0 \end{vmatrix} \]

### Step 4: Expand the determinant
Since the last column contains two zeros, we can expand the determinant along the last column:

\[ D = 1 \cdot \begin{vmatrix} (q - p) \log r & q - p \\ (r - p) \log r & r - p \end{vmatrix} \]

Calculating this 2x2 determinant:

\[ D = (q - p)(r - p) \log r - (r - p)(q - p) \log r = 0 \]

### Conclusion
Thus, the value of the determinant is:

\[ \boxed{0} \]