Question

find 2 numbers whose arithmetic mean exceeds their geometric mean by 2 .and whose harmonic mean is 1/5 of the larger number.....

Answer 1

let the two numbers be x and y , where x is the larger
geometric mean = √(xy)
arithmetic mean = (x+y)/2

(x+y)/2- √xy = 2
√xy = (x+y)/2 - 2
square both sides
xy = (x^2 + 2xy + y^2)/4 - 2(x+y) + 4
times 4
4xy = x^2 + 2xy + y^2 - 8x - 8y + 16
x^2 + y^2 - 2xy - 8x - 8y + 16 = 0
(x-y)^2 - 8(x+y) + 16 = 0

with the help of Wolfram, I found
x = 4n^2, y = 4(n-1)^2 , where n is an integer, as solutions for this
e.g. n = 3 --->x= 36, y= 16
arithmetic mean = (36+16)/2 = 26
geometic mean = √(36x16) = √576 = 24
the arithmetic mean is 2 more than the geometric


The harmonic mean is defined as
the reciprocal of the arithmetic mean of the reciprocals
(had to look that one up)

so arithmetic mean of the reciprocals
= (1/x + 1/y)/2
= (x+y)/(2xy)

so (x+y)/(2xy) = x/5
5x + 5y = 2x^2 y
2x^2 y - 5y = 5x
y(2x^2 - 5) = 5x
y = 5x/(2x^2 - 5)

arrggghhhhh!!!! , sub y into the other equation and solve
Again, I let Wolfram solve this and got

x = 6.86508
y = .384561 , which works in both equations

http://www.wolframalpha.com/input/?i=solve+%28x-y%29%5E2+-+8%28x%2By%29+%2B+16+%3D+0%2C+5x+%2B+5y+%3D+2x%5E2+y

Answer 2

Reinny

In

w o l f r a m a l p h a . com

type:

solve ( x + y ) / 2 = sqroot ( x * y ) + 2 , 2 / ( 1 / x + 1 / y ) = x / 5

The solutions are :

9 and 1