A) 6, 3
B) 4, 27
C) 6, 7
D) 3, 9
Correct Answer: A
Solution :
Given, harmonic mean\[=4\] ...(i) and \[{{G}^{2}}+2A=27\] ...(ii) We know that, \[{{G}^{2}}=H.A\] \[\Rightarrow \] \[{{G}^{2}}=4A\] [from Eq.(i)] \[\therefore \] from Eq. (ii) \[4A+2A=27\] \[\Rightarrow \] \[6A=27\] \[\Rightarrow \] \[A=\frac{27}{6}\] \[\therefore \] \[{{G}^{2}}=4.\frac{27}{6}\] \[\Rightarrow \] \[{{G}^{2}}=18\] Let the numbers are a and b. \[\therefore \] \[\frac{a+b}{2}=\frac{27}{6}\]and\[{{G}^{2}}=ab=18\] \[\Rightarrow \] \[a+b=9\] and \[ab=18\] ...(iii) \[\therefore \] \[{{(a-b)}^{2}}={{(a+b)}^{2}}-4ab\] \[\Rightarrow \] \[a-b=\sqrt{{{9}^{2}}-4\times 18}\] \[\Rightarrow \] \[a-b=\sqrt{9}\] \[\Rightarrow \] \[a-b=3\] ...(iv) From Eqs. (iii) and (iv), we get \[\therefore \] \[a=6,b=3\]
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