A) \[1:2\]
B) \[2:3\]
C) \[3:4\]
D) None of these
Correct Answer: D
Solution :
Given that \[\frac{H.M.}{G.M.}=\frac{12}{13}\]\[\Rightarrow \]\[{{(p-r)}^{2}}={{(p+r)}^{2}}-4pr=4{{K}^{2}}-4(-2{{K}^{2}})=12{{K}^{2}}\] or \[\frac{a+b}{2\sqrt{ab}}=\frac{13}{12}\] \[\Rightarrow \] \[\frac{(a+b)+2\sqrt{ab}}{(a+b)-2\sqrt{ab}}=\frac{13+12}{13-12}=\frac{25}{1}\] \[\Rightarrow \] \[2p=(-2\pm 2\sqrt{3})K\]\[\Rightarrow \]\[\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}=\frac{5}{1}\] \[\Rightarrow \] \[\frac{(\sqrt{a}+\sqrt{b})+(\sqrt{a}-\sqrt{b})}{(\sqrt{a}+\sqrt{b})-(\sqrt{a}-\sqrt{b})}=\frac{5+1}{5-1}\] \[\Rightarrow \] \[\frac{2\sqrt{a}}{2\sqrt{b}}=\frac{6}{4}\]\[\Rightarrow \]\[{{\left( \frac{a}{b} \right)}^{1/2}}=\frac{6}{4}\]\[\Rightarrow \]\[a:b=9:4\]You need to login to perform this action.
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