The harmonic mean of two numbers is 4 and the | vedclass

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(A) Let the two numbers be $x$ and $y$.The arithmetic mean is $A = \frac{x + y}{2}$ and the geometric mean is $G = \sqrt{xy}$.The harmonic mean is giv

Vedclass · Accepted Answer

$6, 3$

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(A) Let the two numbers be $x$ and $y$.The arithmetic mean is $A = \frac{x + y}{2}$ and the geometric mean is $G = \sqrt{xy}$.The harmonic mean is given by $H = \frac{2xy}{x + y} = 4$.From $H = 4$,we have $\frac{2xy}{x + y} = 4$,which implies $xy = 2(x + y)$.Since $A = \frac{x + y}{2}$,we have $x + y = 2A$,so $xy = 2(2A) = 4A$.Since $G^2 = xy$,we have $G^2 = 4A$.Given the relation $2A + G^2 = 27$,substitute $G^2 = 4A$ into the equation:$2A + 4A = 27 \Rightarrow 6A = 27 \Rightarrow A = \frac{27}{6} = 4.5$.Now,$x + y = 2A = 2(4.5) = 9$ and $xy = 4A = 4(4.5) = 18$.The quadratic equation with roots $x$ and $y$ is $t^2 - (x + y)t + xy = 0$,which is $t^2 - 9t + 18 = 0$.Factoring the quadratic: $(t - 6)(t - 3) = 0$.Thus,the numbers are $6$ and $3$.

The harmonic mean of two numbers is $4$ and the arithmetic and geometric means satisfy the relation $2A + G^2 = 27$. The numbers are:

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