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 $p$ and $q$ are two arithmetic means between two numbers and $G$ is the geometric mean between them,then $G^2 = \dots \dots$.

If the $A.M.$ and $H.M.$ of two numbers are $27$ and $12$ respectively,then the $G.M.$ of the two numbers will be:

If $A.M.$ and $G.M.$ of $x$ and $y$ are in the ratio $p : q$,then $x : y$ is

The product of $n$ positive numbers is unity. Their sum is

The sum of two numbers is $6$ times their geometric mean. Show that the numbers are in the ratio $(3+2 \sqrt{2}):(3-2 \sqrt{2})$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module