A) (a) \[2b=30+c\]
B) (b) a=b=c
C) (c) \[{{b}^{2}}=\sqrt{\frac{ac}{8}}\]
D) (d) None of these
Correct Answer: B
Solution :
[b] Since, \[a,b,c\]be in A.P. \[\Rightarrow 2b=a+c\] ,........(1) and \[{{a}^{2}},{{b}^{2}},{{c}^{2}}\]be In H.P. \[{{b}^{2}}=\frac{2{{a}^{2}}{{c}^{2}}}{{{a}^{2}}+{{c}^{2}}}\Rightarrow {{\left( \frac{a+c}{2} \right)}^{2}}=\frac{2{{a}^{2}}{{c}^{2}}}{{{a}^{2}}+{{c}^{2}}}\] \[\Rightarrow ({{a}^{2}}+{{c}^{2}}+2ca)({{a}^{2}}+{{c}^{2}})=8{{a}^{2}}{{c}^{2}}\] \[\Rightarrow {{({{a}^{2}}+{{c}^{2}})}^{2}}+2ca({{a}^{2}}+{{c}^{2}})-8{{a}^{2}}{{c}^{2}}=0\] \[\Rightarrow {{\left( {{a}^{2}}-{{c}^{2}} \right)}^{2}}+2ca\left( {{a}^{2}}-{{c}^{2}} \right)=0\] \[\Rightarrow \left( {{a}^{2}}-{{a}^{2}} \right)\left( {{a}^{2}}-{{c}^{2}}-2ca \right)=0\] If \[\left( {{a}^{2}}-{{c}^{2}} \right)=0\Rightarrow a=c\] From equation (1), we have \[2b=a+c\Rightarrow b=a=c\] Hence, option [b]is correct.
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