H.M. of two numbers is $4$ and their A.M. is $A$ AND G.M. is $G$ ,then

Question

H.M. of two numbers is $4$ , $A$ is A.M. and $G$ IS G.M. then $2A-G^2=27$, the numbers are

$(6, 3)$

$(5,4)$

$(5,-25)$

Answer 1

$$\dfrac{2ab}{a+b}=4$$

$$G^2=ab,2A=a+b$$

$$\implies27=a+b-ab=-\dfrac{ab}2\implies ab=-54$$

$a+b=\dfrac{ab}2=-27$

So, $a,b$ are the roots of

$$t^2-(-27)t-54=0$$

Answer 2

You might know the property that $G^2=AH$

Hence the equation transforms to $$2A-AH=27$$

But $H=4$

Hence $$-2A=27\Rightarrow A=\frac {-27}{2}$$

Which on solving gives $G=i\sqrt {54}$

Hence their exist no real numbers with those properties

If you need to find them it is simple. Let $a$ and $b$ be those numbers.

Hence we have $ab=-54$ and $a+b=-27$

Hence they are roots of the quadratic $$x^2+27x-54=0$$