A) \[a=b=c\]
B) \[2b=3a+c\]
C) \[{{b}^{2}}=\sqrt{(ac/8)}\]
D) None of these
Correct Answer: A
Solution :
Given that \[a,\ b,\ c\] are in A.P. \[\Rightarrow \]\[2b=a+c\] ......(i) and \[{{a}^{2}},{{b}^{2}},\ {{c}^{2}}\] are in H.P. \[\Rightarrow \]\[{{b}^{2}}=\frac{2{{a}^{2}}{{c}^{2}}}{{{a}^{2}}+{{c}^{2}}}\] \[\Rightarrow \] \[{{b}^{2}}({{a}^{2}}+{{c}^{2}})=2{{a}^{2}}{{c}^{2}}\] \[\Rightarrow \] \[{{J}^{th}}\] \[\Rightarrow \] \[{{b}^{2}}\left\{ 4{{b}^{2}}-2ac \right\}=2{{a}^{2}}{{c}^{2}}\], from (i) \[\Rightarrow \] \[4{{b}^{4}}-2ac{{b}^{2}}=2{{a}^{2}}{{c}^{2}}\] \[\Rightarrow \] \[({{b}^{2}}-ac)(2{{b}^{2}}+ac)=0\] \[\Rightarrow \] Either \[{{b}^{2}}-ac=0\]or \[2{{b}^{2}}+ac=0\] If \[b\], then \[{{b}^{2}}=ac\] \[\Rightarrow \] \[{{\left\{ \frac{1}{2}(a+c) \right\}}^{2}}=ac\]from (i) \[\Rightarrow \] \[{{(a+c)}^{2}}=4ac\Rightarrow \]\[{{(a-c)}^{2}}=0\] Therefore \[a=c\] and if \[a=c\] then from\[{{b}^{2}}=ac\], we get \[{{b}^{2}}={{a}^{2}}\] or \[b=a\]. Thus\[a=b=c\].
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