A) \[\frac{A}{G}\]
B) \[\frac{2A}{{{G}^{2}}}\]
C) \[\frac{A}{2{{G}^{2}}}\]
D) \[\frac{A}{{{G}^{2}}}\]
Correct Answer: B
Solution :
Since,\[{{H}_{1}}\]and\[{{H}_{2}}\]are two harmonic means between two positive numbers a and b, then a, \[{{H}_{1}},{{H}_{2}}\]and b are in HP. \[\therefore \] \[{{H}_{1}}=\frac{3ab}{a+2b},{{H}_{2}}=\frac{3ab}{2a+b}\] Now, \[\frac{{{H}_{1}}+{{H}_{2}}}{{{H}_{1}}{{H}_{2}}}=\frac{1}{{{H}_{2}}}+\frac{1}{{{H}_{1}}}\] \[=\frac{2a+b}{3ab}+\frac{a+2b}{3ab}\] \[=\frac{3a+3b}{3ab}=\frac{a+b}{ab}\] ?.(i) Now, A is the arithmetic mean between a and b, then \[2A=a+b\] ?(ii) and G is the geometric mean between a and b, then \[ab={{G}^{2}}\] ...(iii) From Eqs. (i), (ii) and (iii), we get \[\frac{{{H}_{1}}+{{H}_{2}}}{{{H}_{1}}{{H}_{2}}}=\frac{2A}{{{G}^{2}}}\]
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