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

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Question

(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

Vedclass · Accepted Answer

$6, 3$

Vedclass · Answer

(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 $H = \frac{2xy}{x+y} = 4$.Since $A = \frac{x+y}{2}$,we have $x+y = 2A$. Substituting this into the harmonic mean formula: $\frac{2xy}{2A} = 4$ $\Rightarrow \frac{xy}{A} = 4$ $\Rightarrow xy = 4A$.Since $G^2 = xy$,we have $G^2 = 4A$.Given the relation $2A + G^2 = 27$,substitute $G^2 = 4A$:$2A + 4A = 27$ $\Rightarrow 6A = 27$ $\Rightarrow A = \frac{27}{6} = 4.5$.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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