(i) If y is the H.M. between x 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. THANK YOU Share with your friends Share 1 Manbar Singh answered this 1.Since, y is the harmonic mean between x and z, then 2y = 1x + 1z⇒2y = z+xzx⇒y = 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 = 4⇒2aba+b = 4⇒aba+b = 2 .....1Now, arithmetic mean, A = a+b2Geometric mean, G = abNow, 2A + G2 = 27⇒2×a+b2 + ab = 27⇒a + b + ab = 27⇒ab = 27 - a+b .....2Putting the value of ab in 1, we get27-a+ba+b = 2⇒2a+b = 27-a+b⇒a + b = 9 .....3Now, put the value of a+b in 2, we getab = 27 - 9 = 18 ....4Now, a-b2 = a+b2-4ab⇒a-b2 = 92 - 4×18 = 81 - 72 = 9⇒a - 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. 3 View Full Answer