A) \[0\]
B) \[1\]
C) \[-1\]
D) None of these
Correct Answer: A
Solution :
[a] Let A be the 1st term and R the common ration of G.P., then; \[a={{T}_{p}}=A{{R}^{p-1}}\therefore \log a=\log A+(p-1)logR\] Similarly, \[\log b=\log A+(q-1)log\,R\] and \[\log \,c=log\,A+(r-1)log\,R\] \[\therefore \,\,\,\,\Delta =\left| \begin{align} & \log A+(p-1)log\,\,R\,\,p\,\,1 \\ & log\,A+(q-1)log\,\,R\,\,q\,\,\,1 \\ & \log \,A+(r-1)log\,\,R\,\,r\,\,\,\,1 \\ \end{align} \right|\] Split into two determinants and in the first take log A common and in the second take log R common \[\Delta =\log A\left| \begin{matrix} 1 & p & 1 \\ 1 & q & 1 \\ 1 & r & 1 \\ \end{matrix} \right|+\log \,\,R\left| \begin{matrix} p-1 & p & 1 \\ q-1 & q & 1 \\ r-1 & r & 1 \\ \end{matrix} \right|\] Apply \[{{C}_{1}}\to {{C}_{1}}-{{C}_{2}}+{{C}_{3}}\] in the second \[\Delta =0+\log R\left| \begin{matrix} 0 & p & 1 \\ 0 & q & 1 \\ 0 & r & 1 \\ \end{matrix} \right|=0\]
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