A) \[a,b,c\] are in A. P.
B) \[a,b,c\]are in G. P.
C) \[a,b,c\]are in H. P.
D) None of these
Correct Answer: B
Solution :
\[\left| \,\begin{matrix} a & b & a\alpha -b \\ b & c & b\alpha -c \\ 2 & 1 & 0 \\ \end{matrix}\, \right|=0\] Þ \[a[-(b\alpha -c)]-b[-2(b\alpha -c)]+[a\alpha -b)(b-2c)]=0\] Þ\[-ab\alpha +ac+2{{b}^{2}}\alpha -2bc+ab\alpha -2ac\alpha -{{b}^{2}}+2bc=0\] Þ \[ac+2{{b}^{2}}\alpha -2ac\alpha -{{b}^{2}}=0\] Þ \[(ac-{{b}^{2}})-2\alpha (ac-{{b}^{2}})=0\] Þ \[ac-{{b}^{2}}=0\]or \[1-2\alpha =0\] \[\Rightarrow \] \[{{b}^{2}}=ac\]or \[\alpha =\frac{1}{2}\] \[\because \,\,\alpha \ne \frac{1}{2}\] (As given in question) So, \[{{b}^{2}}=ac\] i.e, \[a,b,c\]are in G.P.
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