A) \[\log xyz\]
B) \[(p-1)(q-1)(r-1)\]
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 \[{{T}_{r}}=a{{R}^{r-1}}=z\] \[\Rightarrow \] \[\log x=\log a+(p-1)\log R\] \[\log y=\log a+(q-1)\log R\] and \[\log z=\log a+(r-1)\log R\] \[\therefore \left| \begin{matrix} \log x & p & 1 \\ \log y & q & 1 \\ \log z & r & 1 \\ \end{matrix} \right|=\left| \begin{matrix} \log a+(p-1) & \log R & p\,\,1 \\ \log a+(q-1) & \log R & q\,\,1 \\ \log a+(r-1) & \log R & r\,\,1 \\ \end{matrix} \right|\] \[=\left| \begin{matrix} \log a & p & 1 \\ \log a & q & 1 \\ \log a & r & 1 \\ \end{matrix} \right|+\left| \begin{matrix} (p-1) & \log R & p\,\,\,1 \\ (q-1) & \log R & q\,\,\,1 \\ (r-1) & \log R & r\,\,\,1 \\ \end{matrix} \right|\] \[=\log a\left| \begin{matrix} 1 & p & 1 \\ 1 & q & 1 \\ 1 & r & 1 \\ \end{matrix} \right|+\log R\left| \begin{matrix} p-1 & p-1 & 1 \\ q-1 & q-1 & 1 \\ r-1 & r-1 & 1 \\ \end{matrix} \right|\] \[({{C}_{2}}\to {{C}_{2}}-{{C}_{3}})\] \[=0+0=0\](\[\because \] two columns are identical)
You need to login to perform this action.
You will be redirected in
3 sec
