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# If AM of two numbers be twice their GM then the numbers are in the ratio

SHAIK AASIF AHAMED
7 years ago

Hello student,
Let the numbers be a,b
AM=(a+b)/2
GM= [sqrt{ab}]
Given AM=2GM
So (a+b)/2=2 [sqrt{ab}]
a+b-4 [sqrt{ab}] =0
Divide each term by b we get
(a/b)+1-4 [sqrt{a/b}] =0
([sqrt{a/b}] )2-4( [sqrt{a/b}] )+1=0
on solving quadratic equation we get
[sqrt{a/b}] =2 [sqrt{3}]
So (a/b)=( [2sqrt{3}] )2
(a/b)=7[4sqrt{3}]
Khushi
14 Points
3 years ago

A.T.Q.

a + b / 2 = 2√ab

a + b = 4√ab

Squaring on both sides...

(a + b)² = 16ab ....(i)

a² + b² + 2ab = 16ab

a² + b² + 2ab - 16ab = 0

a² + b² - 14ab = 0

a² + b² - 2ab - 12ab = 0

(a - b)² - 12ab = 0

(a - b)² = 12ab ....(ii)

Dividing (i) by (ii)

(a + b)² / (a - b)² = 16ab / 12ab

(a + b / a - b)² = 4/3

Square root on both sides...

a + b / a - b = 2 /√3

Using componento and dividendo ....

a + b + a - b / a + b - a + b = 2 + √3 / 2 - √3

2a / 2b = 2 + √3 / 2 - √3

a / b = 2 + √3 / 2 - √3

Hence proved.

Hope you can understand it.