A) 0
B) \[1\]
C) \[-1\]
D) None of these
Correct Answer: A
Solution :
As given \[{{b}^{2}}=ac\] and \[\frac{2}{c-a}=\frac{1}{a-b}+\frac{1}{b-c}=\frac{a-c}{(a-b)(b-c)}\] \[\Rightarrow \] \[2(a-b)(b-c)=-{{(a-c)}^{2}}\] \[\Rightarrow \] \[-{{(a-c)}^{2}}=2(ab-2{{b}^{2}}+bc)=2b\left\{ a-2\sqrt{ac}+c \right\}\] \[\Rightarrow \] \[-{{\left\{ (\sqrt{a}-\sqrt{c})(\sqrt{a}+\sqrt{c}) \right\}}^{2}}=2b{{(\sqrt{a}-\sqrt{c})}^{2}}\] Þ\[2b=-\left\{ a+2\sqrt{ac}+c \right\}=-a-2b-c\] or\[a+4b+c=0\].
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