(i)   If y is the  H.M. between and z , prove that, 1/(y-x) + 1/(y-z) = 1/x + 1/z
(ii)
 If harmonic mean of two numbers is 4 , their arithmetic mean 'A' and the geometric mean G satisfy the relation 2A + G2 = 27 . Find the numbers.   [ Answer : 6, 3  ]  

In order to save time I have given two questions in one post. Please give answer of both the questions ans please donot give any link.


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1.Since, y is the harmonic mean between x and z, then     2y = 1x + 1z2y = z+xzxy = 2zxz+xSo, RHS = 1x + 1z LHS = 1y-x + 1y-z=12zxz+x-x + 12zxz+x - z=z+x2zx - xz - x2 + z+x2zx - zx-z2=z+xzx-x2 + z+xzx-z2=z+x1zx-x2 + 1zx-z2=z+xzx-z2+zx-x2zx-x2zx-z2=z+x2zx-x2-z2xz-x . zx-z=z+x-x2+z2-2xz-xzx-z2=z+xx-z2xzx-z2=z+xxz=1x + 1zSo, LHS = RHS2.Let a and b are the 2 numbers.Now, HM between a and b = 42aba+b = 4aba+b = 2    .....1Now, arithmetic mean, A = a+b2Geometric mean, G = abNow, 2A + G2 = 272×a+b2 + ab = 27a + b + ab = 27ab = 27 - a+b    .....2Putting the value of ab in 1, we get27-a+ba+b = 22a+b = 27-a+ba + b = 9    .....3Now, put the value of a+b in 2, we getab = 27 - 9 = 18   ....4Now, a-b2 = a+b2-4aba-b2 = 92 - 4×18 = 81 - 72 = 9a - b = 3   ...5    or     a - b = -3   ....6Adding 3 and 5, we get2a = 12 a = 6Put a = 6 in 3, we get b = 3Adding 3 and 6, we get2a = 6  a = 3Put a = 3 in 3, we get b = 6So, the required numbers are 3 and  6.

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