A) 19
B) 20
C) 21
D) 22
Correct Answer: D
Solution :
As b, a, c are in GP \[GP\Rightarrow {{a}^{2}}=bc\] \[\Rightarrow b+c=abc-a={{a}^{3}}-a\] \[\therefore \]b & c are roots of equation \[{{x}^{2}}-({{a}^{3}}-a)x+{{a}^{2}}=0\] As roots are real, so \[D\ge 0\] \[{{({{a}^{3}}-a)}^{2}}-4{{a}^{2}}\ge 0\Rightarrow {{a}^{2}}\ge 3\] \[\Rightarrow {{a}^{4}}+{{a}^{2}}\Rightarrow 12\]
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