A) \[a\]
B) \[b\]
C) \[c\]
D) None of these
Correct Answer: D
Solution :
As given \[{{b}^{2}}=ac\] and \[\frac{2}{c-a}=\frac{1}{b-c}+\frac{1}{a-b}\] \[\Rightarrow \] \[2(b-c)(a-b)=-{{(a-c)}^{2}}\] \[\Rightarrow \] \[2(ab-ac-{{b}^{2}}+bc)=-{{\left\{ (\sqrt{a}+\sqrt{c})(\sqrt{a}-\sqrt{c}) \right\}}^{2}}\] \[\Rightarrow \] \[2(ab-2{{b}^{2}}+bc)=-{{(\sqrt{a}-\sqrt{c})}^{2}}{{(\sqrt{a}+\sqrt{c})}^{2}}\] \[\Rightarrow \] \[2b{{(\sqrt{a}-\sqrt{c})}^{2}}=-{{(\sqrt{a}-\sqrt{c})}^{2}}{{(\sqrt{a}+\sqrt{c})}^{2}}\] Þ\[2b=-(a+c+2\sqrt{ac}),\ (\because \ \sqrt{a}-\sqrt{c}\ne 0)\]=\[-(a+c+2b)\] \[\Rightarrow \]\[a+b+c=-3b=-3\sqrt{ac}\] is not independent of \[a,\ b\] and \[c\].
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