A) \[6,\,3\]
B) 5, 4
C) \[5,\ -2.5\]
D) \[-3,\ 1\]
Correct Answer: A
Solution :
Let numbers be \[x\] and \[y\]. Then \[A=\frac{1}{2}(x+y),\ \sqrt{xy}=G\] or \[{{G}^{2}}=xy\] and \[\left( \frac{1}{b}+\frac{1}{c}-\frac{1}{a} \right)\left( \frac{1}{c}+\frac{1}{a}-\frac{1}{b} \right)\], \[\Rightarrow \]\[{{G}^{2}}=4A\] Also, \[=\left( \frac{3}{b}-\frac{2}{a} \right)\left( \frac{1}{b} \right)=\frac{3}{{{b}^{2}}}-\frac{2}{ab}\]\[\Rightarrow \]\[(\because \ a,\ b,\ c\] So, \[x+y=9,\ xy=18\] Hence numbers are 6, 3.
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