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The ratio of the A.M. and G.M. of two po...

The ratio of the A.M. and G.M. of two positive numbers a and b, is m : n. Show that a : b = `(m+sqrt(m^2-n^2)):(m-sqrt(m^2-n^2))`.

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To solve the problem, we need to show that if the ratio of the Arithmetic Mean (A.M.) and Geometric Mean (G.M.) of two positive numbers \( a \) and \( b \) is \( m:n \), then the ratio \( a:b \) can be expressed as: \[ \frac{a}{b} = \frac{m + \sqrt{m^2 - n^2}}{m - \sqrt{m^2 - n^2}} \] ### Step-by-Step Solution: ...
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If the ratio of A.M. and G.M. of two positive numbers a and b is m : n, then prove that : a:b=(m+sqrt(m^(2)-n^(2))):(m-sqrt(m^(2)-n^(2)))

If the ratio of A.M.between two positive real numbers a and b to their H.M.is m:n then a:b is equal to

Knowledge Check

  • If the A.M of two positive numbers a and b, (a gt b) is twice their G.M. , then a:b is :

    A
    a)`2: sqrt3`
    B
    b)`2:7 + 4sqrt3`
    C
    c)`2+ sqrt3 : 2-sqrt3`
    D
    d)`7+4sqrt3 :7-4sqrt3`

  • If the A.M. of two positive numbers a and b, (agtb) , is twice their G.M., then a:b is :

    A
    `2:sqrt3`
    B
    `2:7+4sqrt3`
    C
    `2+sqrt3:2-sqrt3`
    D
    `7+4sqrt3:7-4sqrt3`

  • The A.M., G.M. and H.M. between two positive numbers a and b are equal, then

    A
    a = b
    B
    ab = 1
    C
    `a gt b`
    D
    `a lt b`

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    The A.M and G.M of two numbers a and b are 18 and 6 respectively then H.M of a and b is

    If sum of two numbers a&b is n xx their G.M then show that (a)/(b)=(n+sqrt(n^(2)-4))/(n-sqrt(n^(2)-4))

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    If A.M.and G.M.between two numbers is in the ratio m:n then prove that the numbers are in the ratio (m+sqrt(m^(2)-n^(2))):(m-sqrt(m^(2)-n^(2)))

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