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Is there any relation between AM, GM, HM in Progressions?

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Qualified (Engg from NIT) & experienced mentor with 6+ yrs experience.

AM×HM=SQR of GM i.e. AM×HM= GM x GM
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Math magician

Yes relationship is present and also equation is also present
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Tutor

GM^2 = AM *HM ,Geometric progression of any two numbers is always a sqaure root product of Arithmetic and Harmonic progressions of those two numbers.A simple googling would have fetched you the answer .Anyways good luck !
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Go hard or Go home

AM*HM=GM*GM
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There are 2 relations between them, 1. A.M. > G.M. > H.M 2. GM^2=AM×HM
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Mathematics Teacher

yes. (GM)^2 =AM* HM
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Tutor

If a and b are two numbers such that a,b>0 , then AM of the numbers = (a+b)/2 GM=(ab)^1/2 HM=2/ = 2ab/(a+b) Thus multiplying AM with HM we get AM*HM = * = ab = ^2 = ^2 Thus we get the co-relation as below; GM = ^1/2
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Dedicated Teacher With Lots of Teaching Experience In Maths For All Boards

AM X HM = GM2(GM square)
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Mathematics And Physics Tutor

A.M of two numbers a and b is (a+b)/2 G.M of two numbers is square root over ab H.M is 2ab/(a+b) If we multiply A.M and H.M then (a+b)/2*2ab/(a+b)=ab=G.M^2 Therefore A.M * H.M = g.M^2 That is A.M,g.M, and H.M are in G.P
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Expert in Maths

GM2 = AM.HM
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