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        If the arithmetic mean of two positive numbers a and b (a>b) is twice their geometric mean , then find the ratio a:b.
one year ago

## Answers : (2)

Arun
22987 Points
							Dear Ankesh [4sqrt{3}](a/b)=72sqrt{3}] )[sqrt{3}]So (a/b)=( [2-4( [sqrt{a/b}] )+1=0on solving quadratic equation we get[sqrt{a/b}] =2 2Please find the answer to your question belowLet the numbers be a,bAM=(a+b)/2GM= [sqrt{ab}]Given AM=2GMSo (a+b)/2=2 [sqrt{ab}]a+b-4 [sqrt{ab}] =0Divide each term by b we get(a/b)+1-4 [sqrt{a/b}] =0([sqrt{a/b}] )  RegardsArun (askIITians forum expert)

one year ago
Arun
22987 Points
							Dear Ankesh arithmetic mean of a and b = (a+b)/2 geometric mean of a and b = √(ab) so, (a+b)/2 = 2√(ab) a + b = 4√(ab) (a+b)^2 = 16ab a^2 + 2ab + b^2 = 16 ab or a^2 - 14ab + b^2 = 0 By the quadratic equation, a = [14b ± √(196b^2 - 4b^2)] / 2 or a = 7b ± 4√3 b or a/b = 7 ± 4√3 and (2+√3)/(2-√3) = 7 + 4√3 { Here we see that the hypothesis is true only when a is the larger of the two numbers a and b. When b is the larger number, 7-4√3 is the ratio. This is the reciprocal of 7 + 4√3, explaining the other solution to the quadratic. i.e., if a/b=7 + 4√3, then b/a=7-4√3 }  RegardsArun (askIITians forum expert)

one year ago
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