If the ratio of A.M. and G.M. between two numbers a and b is m: n, pro

1.	If the ratio of A.M. and G.M. between two numbers a and b is m: n, prove that2 2.m -n
Answer» Let the two numbers be a and b, AM = a+b/2 and GM = √(ab) So ….(a+b/2)/√(ab) = m/n, Applying componendo and dividendo i.e if p/q = l/m then p+q/p-q = l+m/l-m, => (a+b+2√(ab))/(a+b-2√(ab)) = m+n/m-n => (√(a)+√(b))^2/(√(a)-√(b))^2 = m+n/m-n => √(a)+√(b)/√(a)-√(b) = √(m+n)/√(m-n) Applying componendo and dividendo, => 2√(a)/2√(b) = √(m+n)+√(m-n)/√(m+n)-√(m-n) Squaring both sides, a/b = (m+n+m-n+2√(m^2-n^2))/(m+n+m-n-2√(m^2-n^2)) => a/b = m+√(m^2-n^2)/m-√(m^2-n^2) hit like if you find it useful please copy use

If the ratio of A.M. and G.M. between two numbers a and b is m: n, prove that2 2.m -n

Answer»

Let the two numbers be a and b,

AM = a+b/2 and GM = √(ab)

So ….(a+b/2)/√(ab) = m/n,

Applying componendo and dividendo i.e if p/q = l/m then p+q/p-q = l+m/l-m,

=> (a+b+2√(ab))/(a+b-2√(ab)) = m+n/m-n

=> (√(a)+√(b))^2/(√(a)-√(b))^2 = m+n/m-n

=> √(a)+√(b)/√(a)-√(b) = √(m+n)/√(m-n)

Applying componendo and dividendo,

=> 2√(a)/2√(b) = √(m+n)+√(m-n)/√(m+n)-√(m-n)

Squaring both sides,

a/b = (m+n+m-n+2√(m^2-n^2))/(m+n+m-n-2√(m^2-n^2))

=> a/b = m+√(m^2-n^2)/m-√(m^2-n^2)

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