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Step-by-Step Solution
Step 1: Express the G.P. terms
Since $l, m, n$ are respectively the $p^{\text{th}}$, $q^{\text{th}}$, and $r^{\text{th}}$ terms of a geometric progression (G.P.) with first term $A$ and common ratio $R$, we can write:
$l = A R^{p-1}, \quad m = A R^{q-1}, \quad n = A R^{r-1}.$
Step 2: Take logarithms
Applying the logarithm to each, we obtain:
$\log l = \log A + (p-1)\log R,$
$\log m = \log A + (q-1)\log R,$
$\log n = \log A + (r-1)\log R.$
Step 3: Form the 3×3 determinant
We need to evaluate:
$
\begin{vmatrix}
\log l & p & 1 \\
\log m & q & 1 \\
\log n & r & 1
\end{vmatrix}.
$
Substitute the expressions for $\log l, \log m, \log n$ to get:
$
\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 4: Perform column operations
A common approach to simplify this determinant is to use column operations. For instance, one may perform:
• New $C_1 = C_1 - (\log R)\,C_2 + (\log R - \log A)\,C_3.$
These operations are designed to factor out and eliminate the terms involving $\log A$ and $\log R$ from the first column.
After applying the above operation carefully, the first column becomes all zeros:
$
\begin{vmatrix}
0 & p & 1 \\
0 & q & 1 \\
0 & r & 1
\end{vmatrix},
$
because each entry in the first column is transformed to:
$
\log A + (p-1) \log R - p\,\log R + \ldots
$
which cancels out to $0$, similarly for the other rows.
Step 5: Conclude the value of the determinant
When the first column of a 3×3 determinant is all zero, the determinant is:
$
\begin{vmatrix}
0 & p & 1 \\
0 & q & 1 \\
0 & r & 1
\end{vmatrix}
= 0.
$
Therefore, the value of the given determinant is 0.
