The harmonic mean of two numbers is 4. If the | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let the two numbers be $x$ and $y$.Given that the harmonic mean $H = 4$,we have $H = \frac{2xy}{x+y} = 4$,which implies $xy = 2(x+y)$.Since $A = \

Vedclass · Accepted Answer

$6, 3$

Vedclass · Answer

(A) Let the two numbers be $x$ and $y$.Given that the harmonic mean $H = 4$,we have $H = \frac{2xy}{x+y} = 4$,which implies $xy = 2(x+y)$.Since $A = \frac{x+y}{2}$,we have $x+y = 2A$,so $xy = 2(2A) = 4A$.We know that $G^2 = xy$,therefore $G^2 = 4A$.Substitute $G^2 = 4A$ into the given equation $2A + G^2 = 27$:$2A + 4A = 27$ $\Rightarrow 6A = 27$ $\Rightarrow A = 4.5$.Since $A = \frac{x+y}{2} = 4.5$,we get $x+y = 9$.Since $xy = 4A = 4(4.5) = 18$,we have the system $x+y = 9$ and $xy = 18$.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: $(t-6)(t-3) = 0$,so $t = 6$ or $t = 3$.Thus,the two numbers are $6$ and $3$.

The harmonic mean of two numbers is $4$. If their arithmetic mean $A$ and geometric mean $G$ satisfy the equation $2A + G^2 = 27$,find the two numbers.

Similar Questions

If $a, b, c$ are in $A.P.$ and $a^2, b^2, c^2$ are in $H.P.$,then

If the Arithmetic Mean of $a$ and $b$ is $\frac{a^{n+1} + b^{n+1}}{a^n + b^n}$,then $n = \dots$

Let $0 < z < y < x$ be three real numbers such that $\frac{1}{x}, \frac{1}{y}, \frac{1}{z}$ are in an arithmetic progression and $x, \sqrt{2}y, z$ are in a geometric progression. If $xy + yz + zx = \frac{3}{\sqrt{2}} xyz$,then $3(x + y + z)^2$ is equal to $............$.

If $a, b, c$ are in $A.P.$ and $b, c, d$ are in $H.P.$,then

If the geometric mean of two positive numbers is $6$ and their arithmetic mean is $6.5$,then the numbers are.........

Vedclass Test Series

Exam Paper Generator

Online Exam Module