Key Concepts and Formulas

• Geometric Progression (G.P.): For a G.P. $a_1, a_2, \ldots, a_{10}$ with first term $A$ and common ratio $R$, the $i$-th term is $a_i = A \cdot R^{i-1}$. Given $a_i > 0$, we have $A > 0$ and $R > 0$.
• Logarithm Properties: $\log_e(xy) = \log_e x + \log_e y$ and $\log_e(x^p) = p \log_e x$.
• Logarithms of G.P. Terms: If $a_1, a_2, \ldots, a_{10}$ are in G.P., then $b_i = \log_e a_i$ form an Arithmetic Progression (A.P.) with first term $b_1 = \log_e a_1$ and common difference $d = \log_e R$. Thus, $b_i = b_1 + (i-1)d$.
• Determinant Properties: A determinant is zero if two of its rows or columns are identical. Applying row/column operations of the form $R_i \to R_i + k R_j$ or $C_i \to C_i + k C_j$ does not change the determinant's value. If the elements of a row (or column) are in A.P., specific column (or row) operations can be used to create identical columns (or rows).

Step-by-Step Solution

Step 1: Simplify the general term of the determinant. Each element in the determinant is of the form $\log_e(a_i^r a_j^k)$. Using logarithm properties, we can rewrite this as: $\log_e(a_i^r a_j^k) = \log_e(a_i^r) + \log_e(a_j^k) = r \log_e a_i + k \log_e a_j$

Step 2: Express the logarithmic terms using the A.P. property. Since $a_1, a_2, \ldots, a_{10}$ are in G.P., the sequence $b_i = \log_e a_i$ forms an A.P. with first term $b_1 = \log_e a_1$ and common difference $d = \log_e R$. Thus, $b_i = b_1 + (i-1)d$. Substituting this into the expression from Step 1: $r b_i + k b_j = r(b_1 + (i-1)d) + k(b_1 + (j-1)d)$ $= (r+k)b_1 + (r(i-1) + k(j-1))d$

Step 3: Analyze the structure of the determinant. Let the determinant be denoted by $\Delta$. The elements of the determinant are: \Delta = \left| {\matrix{ {{r \log_e a_1 + k \log_e a_2}} & {{r \log_e a_2 + k \log_e a_3}} & {{r \log_e a_3 + k \log_e a_4}} \cr {{r \log_e a_4 + k \log_e a_5}} & {{r \log_e a_5 + k \log_e a_6}} & {{r \log_e a_6 + k \log_e a_7}} \cr {{r \log_e a_7 + k \log_e a_8}} & {{r \log_e a_8 + k \log_e a_9}} & {{r \log_e a_9 + k \log_e a_{10}}} \cr } } \right| Using $b_i = \log_e a_i$, the elements are: \Delta = \left| {\matrix{ {r b_1 + k b_2} & {r b_2 + k b_3} & {r b_3 + k b_4} \cr {r b_4 + k b_5} & {r b_5 + k b_6} & {r b_6 + k b_7} \cr {r b_7 + k b_8} & {r b_8 + k b_9} & {r b_9 + k b_{10}} \cr } } \right| Let's examine the first row. The difference between consecutive terms is: $(r b_2 + k b_3) - (r b_1 + k b_2) = r(b_2 - b_1) + k(b_3 - b_2) = rd + kd = (r+k)d$. Similarly, $(r b_3 + k b_4) - (r b_2 + k b_3) = r(b_3 - b_2) + k(b_4 - b_3) = rd + kd = (r+k)d$. Thus, the first row is an A.P. with common difference $(r+k)d$.

Let's check the second row: $(r b_5 + k b_6) - (r b_4 + k b_5) = r(b_5 - b_4) + k(b_6 - b_5) = rd + kd = (r+k)d$. And $(r b_6 + k b_7) - (r b_5 + k b_6) = r(b_6 - b_5) + k(b_7 - b_6) = rd + kd = (r+k)d$. The second row is also an A.P. with common difference $(r+k)d$.

Similarly, the third row is an A.P. with common difference $(r+k)d$. Let $X_{m1}$ be the first element of row $m$, and let $D = (r+k)d$. Then the determinant can be written as: $\Delta = \begin{vmatrix} X_{11} & X_{11}+D & X_{11}+2D \\ X_{21} & X_{21}+D & X_{21}+2D \\ X_{31} & X_{31}+D & X_{31}+2D \end{vmatrix}$

Step 4: Simplify the determinant using column operations. We apply the column operations $C_2 \to C_2 - C_1$ and $C_3 \to C_3 - C_1$. The new determinant is: $\Delta = \begin{vmatrix} X_{11} & D & 2D \\ X_{21} & D & 2D \\ X_{31} & D & 2D \end{vmatrix}$ Now, we can factor out $D$ from the second column and $2D$ from the third column: $\Delta = D \cdot (2D) \begin{vmatrix} X_{11} & 1 & 1 \\ X_{21} & 1 & 1 \\ X_{31} & 1 & 1 \end{vmatrix}$ Since the second and third columns are identical, the value of the determinant is 0. Therefore, $\Delta = 0$ for all $r, k \in \mathbb{N}$.

Step 5: Determine the number of elements in S. The set $S$ consists of pairs $(r, k)$ where $r, k \in \mathbb{N}$ and the determinant is 0. Since the determinant is always 0 for any natural numbers $r$ and $k$, every pair of natural numbers $(r, k)$ satisfies the condition. The set of natural numbers $\mathbb{N}$ is infinite. Thus, there are infinitely many such pairs $(r, k)$.

Common Mistakes & Tips

• Logarithm-A.P. Connection: Recognize that a G.P. becomes an A.P. when logarithms are applied. This is a key simplification tool.
• Determinant Structure: Look for patterns like A.P.s in rows or columns, which allow for simplification using column/row operations to create identical columns/rows, leading to a determinant of zero.
• Case $d=0$: If the common ratio $R=1$, then $d=\log_e 1 = 0$. In this case, $(r+k)d = 0$. The determinant still evaluates to 0, so this case does not change the outcome of the determinant being zero.

Summary

The problem involves a determinant whose entries are logarithmic expressions of terms in a geometric progression. By utilizing the property that the logarithms of terms in a G.P. form an A.P., we expressed the determinant's elements in terms of the first term of the A.P. and its common difference. We then observed that each row of the determinant represented terms in an A.P. with the same common difference. Applying column operations based on this A.P. structure led to a determinant with two identical columns, making its value identically zero. Since the determinant is always zero for any natural numbers $r$ and $k$, the set $S$ contains all possible pairs of natural numbers, which is an infinite set.

The final answer is $\boxed{infinitely many}$.