A) \[pqr\]
B) 0
C) \[p+q+r\]
D) \[pq+qr+rp\]
E) \[{{(p+q+r)}^{2}}\]
Correct Answer: B
Solution :
\[\because \] a, b, c are the p th, q th and r th terms of GP \[\therefore \] \[a=A{{R}^{p-1}}\] \[\Rightarrow \] \[loga=logA+(p-1)\log R\] Similarly, \[b=A{{R}^{q-1}}\] \[\Rightarrow \] \[log\text{ }b=log\text{ }A+(q-1)log\text{ }R\] and \[c=A{{R}^{r-1}}\] \[\Rightarrow \] \[log\text{ }c=log\text{ }A+(r-1)log\text{ }R\] Now, \[\left| \begin{matrix} {{\log }_{e}}a & p & 1 \\ {{\log }_{e}}b & q & 1 \\ {{\log }_{e}}c & r & 1 \\ \end{matrix} \right|=(q-r)\log a+\] \[(r-p)logb+\text{(}p-q)logc\] \[=(q-r+r-p+p-q)log\text{ }A+\{(q-r)\}\] \[(p-1)+(r-p)(q-1)+(p-q)(r-1)\}\] \[log\text{ }R=0.log\text{ }A+0.log\text{ }R=0\]
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