• # question_answer The harmonic mean of two numbers is 4 and the arithmetic and geometric means satisfy the relation $2A+{{G}^{2}}=27$, the numbers are                   [MNR 1987; UPSEAT 1999, 2000] A) $6,\,3$ B) 5, 4 C) $5,\ -2.5$ D) $-3,\ 1$

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.