A) \[A=G\]
B) \[A>G\]
C) \[A<G\]
D) \[A=-\,G\]
Correct Answer: B
Solution :
\[{{x}^{2}}-2Ax+{{G}^{2}}=0\] ?..(i) Let \[a,b\] are the two numbers whose A.M. is A and G.M. is G. \[\therefore \]\[A=(a+b)/2,{{G}^{2}}=ab\] From (i), \[{{x}^{2}}-(a+b)x+ab=0\] Þ \[{{x}^{2}}-ax-bx+ab=0\]\[\Rightarrow x(x-a)-b(x-a)=0\] Þ \[(x-a)(x-b)=0\] \[\therefore a,\,b\] are the roots of the equation \[\therefore \] \[A-G=\frac{a+b}{2}-\sqrt{ab}\]\[=\frac{1}{2}[(a+b)-2\sqrt{ab}]\] \[=\frac{1}{2}{{(\sqrt{a}-\sqrt{b})}^{2}}>0\Rightarrow \]\[A>G\].
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