l, m,n are the `p^(th), q ^(th) and r ^(th)` term of a G.P. all positive, then `|{:(logl, p, 1),(log m, q, 1),(log n ,r,1):}|` equals :

Options

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

Answer

To solve the problem, we need to find the value of the determinant given the terms of a geometric progression (G.P.). Let's denote the terms of the G.P. as follows:

1. Let \( l \) be the \( p^{th} \) term of the G.P.
2. Let \( m \) be the \( q^{th} \) term of the G.P.
3. Let \( n \) be the \( r^{th} \) term of the G.P.

### Step 1: Express the terms of the G.P.
The \( n^{th} \) term of a G.P. can be expressed as:
\[ T_n = a \cdot r^{n-1} \]
where \( a \) is the first term and \( r \) is the common ratio.

Thus, we can write:
- \( l = a \cdot r^{p-1} \)
- \( m = a \cdot r^{q-1} \)
- \( n = a \cdot r^{r-1} \)

### Step 2: Take logarithms
Now, we take the logarithm of each term:
\[ \log l = \log(a \cdot r^{p-1}) = \log a + (p-1) \log r \]
\[ \log m = \log(a \cdot r^{q-1}) = \log a + (q-1) \log r \]
\[ \log n = \log(a \cdot r^{r-1}) = \log a + (r-1) \log r \]

### Step 3: Set up the determinant
We need to find the determinant:
\[ D = \begin{vmatrix} \log l & p & 1 \\ \log m & q & 1 \\ \log n & r & 1 \end{vmatrix} \]

### Step 4: Substitute the logarithmic values into the determinant
Substituting the expressions for \( \log l \), \( \log m \), and \( \log n \):
\[ D = \begin{vmatrix} \log a + (p-1) \log r & p & 1 \\ \log a + (q-1) \log r & q & 1 \\ \log a + (r-1) \log r & r & 1 \end{vmatrix} \]

### Step 5: Simplify the determinant
We can simplify this determinant by performing row operations. 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 6: Factor out common terms
Notice that the last column has two zeros, which means we can factor out \( \log r \) from the second and third rows:
\[ D = \log r \cdot \begin{vmatrix} \log a + (p-1) \log r & p & 1 \\ q - p & 1 & 0 \\ r - p & 1 & 0 \end{vmatrix} \]

### Step 7: Evaluate the determinant
Now, we can evaluate the determinant:
\[ D = \log r \cdot \begin{vmatrix} \log a + (p-1) \log r & p \\ q - p & 1 \\ r - p & 1 \end{vmatrix} \]
Since the second and third rows are identical, the determinant evaluates to zero:
\[ D = 0 \]

### Final Answer
Thus, the value of the determinant is:
\[ |{:(\log l, p, 1), (\log m, q, 1), (\log n, r, 1):}| = 0 \]