If the A.M. of two positive numbers `aa n db(a > b)` is twice their geometric mean. Prove that : `a : b=(2+sqrt(3)):(2-sqrt(3))dot`

Explanation

Let A be the A.M. and G be the G.M. between a and b .
then ` A = (a + b)/(2) and G = sqrt(ab)`
Given, ` A = 2G `
`rArr (a+b)/(2) = 2sqrt(ab) `
`rArr (a + b)/(2sqrt(ab)) = (2)/(1)`
`rArr (a + b + 2sqrt(ab))/(a + b - 2 sqrt(ab)) = (3)/(1)` " " [By componendo and dividendo]
`rArr ((sqrt(a) + sqrt(b))^(2))/((sqrt(a) - sqrt(b))^(2)) = (3)/(1)`
`rArr (sqrt(a) + sqrt(b))/(sqrt(a) - sqrt(b))=(sqrt(3))/(1)`
`rArr (2sqrt(a))/(2sqrt(b)) = ((sqrt(3)+1)/(sqrt(3)-1))` " " [By componendo and dividendo ]
`rArr (a)/(b) ((sqrt(3) +1)/(sqrt(3) -1))^(2) = (4 + 2sqrt(4))/(4-2sqrt(3))`
`rArr (a)/(b) = (2+ sqrt(3))/(2-sqrt(3))` Hence proved .