A) \[\log xyz\]
B) \[(p-q)\,(q-1)\,(r-1)\]
C) \[pqr\]
D) \[0\]
Correct Answer: D
Solution :
Let a and R be the first term and common ratio of a GP. \[\therefore \] \[{{T}_{p}}=a{{R}^{p-1}}=x\] \[{{T}_{q}}=a{{R}^{q-1}}=y\] and \[{{T}_{r}}=a{{R}^{r-1}}=z\] \[\text{log }x=\text{log }a+\left( p-1 \right)\,\,\text{log }R\] \[\text{log }y=\text{log }a+\left( q-1 \right)\,\,\text{log }R\] and \[\text{log }z=\text{log }a+\left( r-1 \right)\,\,\text{log }R\] \[\therefore \left| \begin{matrix} \log x & p & 1 \\ \log y & q & 1 \\ \log z & r & 1 \\ \end{matrix} \right|=\left| \begin{matrix} \log a+(p-1)\,\log R & p & 1 \\ \log a+(q-1)\log R & q & 1 \\ \log a+(r-1)\,\log R & r & 1 \\ \end{matrix} \right|\] \[=\left| \begin{matrix} \log a & p & 1 \\ \log a & q & 1 \\ \log a & r & 1 \\ \end{matrix} \right|+\left| \begin{matrix} (p-1)\,\log R & p & 1 \\ (q-1)\log R & q & 1 \\ (r-1)\,\log R & r & 1 \\ \end{matrix} \right|\] \[=\log a\left| \begin{matrix} 1 & p & 1 \\ 1 & q & 1 \\ 1 & r & 1 \\ \end{matrix} \right|+\log R\left| \begin{matrix} p-1 & p-1 & 1 \\ q-1 & q-1 & 1 \\ r-1 & r & 1 \\ \end{matrix} \right|\] \[({{C}_{2}}\to {{C}_{2}}-{{C}_{3}})\] \[=0+0=0\] (\[\because \] two columns are identical)
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