Let a_1, a_2, a_3, dots, a_{10} be in G.P. wi | vedclass

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(B) Let the $G.P.$ be $a_n = a \cdot x^{n-1}$,where $a > 0$ and $x > 0$.Then $\log_e(a_n^r a_{n+1}^k) = r \log_e(a_n) + k \log_e(a_{n+1}) = r \log_e(a

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infinitely many

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(B) Let the $G.P.$ be $a_n = a \cdot x^{n-1}$,where $a > 0$ and $x > 0$.Then $\log_e(a_n^r a_{n+1}^k) = r \log_e(a_n) + k \log_e(a_{n+1}) = r \log_e(a \cdot x^{n-1}) + k \log_e(a \cdot x^n) = r(\log_e a + (n-1)\log_e x) + k(\log_e a + n \log_e x) = (r+k)\log_e a + (r(n-1) + kn)\log_e x$.Let $A = \log_e a$ and $B = \log_e x$. Then the term is $(r+k)A + (r(n-1) + kn)B$.Notice that the terms in the determinant are linear expressions in $n$. Specifically,let $f(n) = c_1 n + c_2$. Since each row of the determinant is an arithmetic progression (as $n$ increases by $1$,the value changes by a constant $rB + kB$),the rows are linearly dependent.Specifically,$R_2 - R_1 = R_3 - R_2$,which implies $R_1 - 2R_2 + R_3 = 0$.Since the rows are in arithmetic progression,the determinant is always $0$ for any $r, k \in N$.Therefore,the set $S$ contains all pairs $(r, k)$ where $r, k \in N$,which means there are infinitely many elements in $S$.

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