Key Concepts and Formulas

1. Geometric Progression (G.P.): The $n^{th}$ term of a G.P. with first term $A$ and common ratio $R$ is $T_n = A R^{n-1}$.
2. Logarithm Properties:
   • $\log(xy) = \log x + \log y$ (Product Rule)
   • $\log(x^y) = y \log x$ (Power Rule)
3. Determinant Properties: If a column (or row) of a determinant consists entirely of zeros, the value of the determinant is zero. Column operations of the form $C_i \to C_i + k C_j$ do not change the value of the determinant.

Step-by-Step Solution

Step 1: Express the G.P. terms using the general formula. Let the first term of the G.P. be $A$ and the common ratio be $R$. We are given that $l, m, n$ are the $p^{th}, q^{th}, r^{th}$ terms, respectively. So, we can write:

• $l = A R^{p-1}$
• $m = A R^{q-1}$
• $n = A R^{r-1}$ We are also given that all terms are positive, which implies $A > 0$ and $R > 0$.

Why this step? This step translates the problem statement into algebraic expressions that we can manipulate. It establishes the relationship between the terms ($l, m, n$) and their positions ($p, q, r$) within the G.P.

Step 2: Apply logarithms to the G.P. terms. The determinant involves $\log l, \log m, \log n$. We apply the logarithm (with any convenient base) to the expressions from Step 1. Using the logarithm properties:

• $\log l = \log(A R^{p-1}) = \log A + \log(R^{p-1}) = \log A + (p-1)\log R$
• $\log m = \log(A R^{q-1}) = \log A + \log(R^{q-1}) = \log A + (q-1)\log R$
• $\log n = \log(A R^{r-1}) = \log A + \log(R^{r-1}) = \log A + (r-1)\log R$

Why this step? Taking logarithms converts the multiplicative structure of the G.P. terms into an additive structure. This is crucial because additive structures are more amenable to simplification using linear operations, such as those available for determinants. Notice how the exponents $(p-1), (q-1), (r-1)$ have become multipliers.

Step 3: Substitute the logarithmic expressions into the determinant. Now, we substitute these expressions into the given determinant: \Delta = \left| {\matrix{ {\log \,l} & p & 1 \cr {\log \,m} & q & 1 \cr {\log \,n} & r & 1 \cr } } \right| = \left| {\matrix{ {\log A + (p-1)\log R} & p & 1 \cr {\log A + (q-1)\log R} & q & 1 \cr {\log A + (r-1)\log R} & r & 1 \cr } } \right|

Why this step? This step sets up the determinant for simplification by replacing the terms $\log l, \log m, \log n$ with their derived algebraic forms.

Step 4: Simplify the determinant using column operations. We can rewrite the elements in the first column ($C_1$) as $\log A - \log R + p \log R$, $\log A - \log R + q \log R$, and $\log A - \log R + r \log R$. Let $X = \log A - \log R$ and $Y = \log R$. Then the first column elements are $X + pY$, $X + qY$, $X + rY$. The determinant is: \Delta = \left| {\matrix{ {X + pY} & p & 1 \cr {X + qY} & q & 1 \cr {X + rY} & r & 1 \cr } } \right| We perform the following column operations:

Operation 1: $C_1 \to C_1 - Y C_2$ This operation subtracts $Y$ times the elements of the second column ($C_2$) from the corresponding elements of the first column ($C_1$).

• Row 1: $(X + pY) - Y(p) = X$
• Row 2: $(X + qY) - Y(q) = X$
• Row 3: $(X + rY) - Y(r) = X$ The determinant becomes: \Delta = \left| {\matrix{ X & p & 1 \cr X & q & 1 \cr X & r & 1 \cr } } \right|

Why this step? This operation aims to isolate the constant part ($\log A - \log R$) in the first column by eliminating the terms dependent on $p, q, r$.

Operation 2: $C_1 \to C_1 - X C_3$ This operation subtracts $X$ times the elements of the third column ($C_3$) from the corresponding elements of the first column ($C_1$).

• Row 1: $X - X(1) = 0$
• Row 2: $X - X(1) = 0$
• Row 3: $X - X(1) = 0$ The determinant becomes: \Delta = \left| {\matrix{ 0 & p & 1 \cr 0 & q & 1 \cr 0 & r & 1 \cr } } \right|

Why this step? Now that the first column contains identical elements ($X$), we can use the third column (which consists of all 1s) to make the entire first column zero.

Step 5: Evaluate the determinant. A determinant with a column of all zeros has a value of zero. $\Delta = 0$

Why this step? This is a fundamental property of determinants. The presence of a zero column implies that the rows (or columns) are linearly dependent in a way that results in a zero determinant.

The value of the determinant is 0.

The correct answer is (D).

Common Mistakes & Tips

• Logarithm Properties: Ensure a firm grasp of $\log(ab) = \log a + \log b$ and $\log(a^b) = b \log a$. Misapplication of these rules is a common error.
• Determinant Operations: Be systematic with column (or row) operations. Aim to create zeros strategically. Remember that operations like $C_i \to C_i + k C_j$ preserve the determinant's value.
• Algebraic Errors: Small mistakes in algebraic manipulation, especially with signs or distributing terms, can lead to an incorrect final answer. Double-check each step.

Summary The problem leverages the properties of Geometric Progressions and Logarithms to transform the terms into a linear form. By taking logarithms of the G.P. terms, we converted the multiplicative relationships into additive ones. These additive expressions were then substituted into the determinant. Through a series of strategic column operations, we successfully reduced the first column of the determinant to all zeros. A fundamental property of determinants states that if any column (or row) is entirely composed of zeros, the determinant's value is zero. Thus, the determinant evaluates to 0.

The final answer is $\boxed{0}$.