Finding $x$ and $y$ Given $A, G, H \in N$ and $A+G+H=49$

[Albert1]

we are given :$x,y\in R ,and \,\, x>y>0, A=\dfrac {x+y}{2},G=\sqrt {xy}, H=\dfrac {2xy}{x+y}$

if $A,G,H\in N, \,\, and \,\,\, A+G+H=49$

please find $,x,y$

 

[Albert1]

we are given :$x,y\in R ,and \,\, x>y>0, A=\dfrac {x+y}{2},G=\sqrt {xy}, H=\dfrac {2xy}{x+y}$

if $A,G,H\in N, \,\, and \,\,\, A+G+H=49$

please find $,x,y$

hint :

Spoiler

$G^2=A\times H$
$A,G,H $ is a geometric sequence
and $A>G>H$

 

[kaliprasad]

hint :

Spoiler

$G^2=A\times H$
$A,G,H $ is a geometric sequence
and $A>G>H$

I hope that you could have given some time before a hint. One day is too short. Here is the solution

Spoiler

Let common ratio be r
so $H + Hr + Hr^2= 49$
so $H(1+r+r^2) = 49$
r need not be integer it could be $\dfrac{p}{q}$ but let us try integer
so we get
H is a factor of 49 so H = 1 or 7
H =1 gives r irrational and H= 7 gives r = 2

so $AM = 28$ or $x+ y = 56$ and $xy =196$

solving these 2 as x is greater $x= 28+14\sqrt{3}$ and $y= 28- 14\sqrt{3}$
For r to be fraction h has to be a product of a sqaure that multyiple has to be factor of 49. So h has to be a square or 7 * square
trying $H = 9$ we get $G = 15$ and $A = 25$ and so $x = 45,y = 5$ as the second solution

 

[Albert1]

I hope that you could have given some time before a hint. One day is too short. Here is the solution

Spoiler

Let common ratio be r
so $H + Hr + Hr^2= 49$
so $H(1+r+r^2) = 49$
r need not be integer it could be $\dfrac{p}{q}$ but let us try integer
so we get
H is a factor of 49 so H = 1 or 7
H =1 gives r irrational and H= 7 gives r = 2

so $AM = 28$ or $x+ y = 56$ and $xy =196$

solving these 2 as x is greater $x= 28+14\sqrt{3}$ and $y= 28- 14\sqrt{3}$
For r to be fraction h has to be a product of a sqaure that multyiple has to be factor of 49. So h has to be a square or 7 * square
trying $H = 9$ we get $G = 15$ and $A = 25$ and so $x = 45,y = 5$ as the second solution

very good !