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Ex 9.3, 29 - If A and G be AM, GM between two positive numbers - AM and GM (Arithmetic Mean And Geometric mean)

  1. Chapter 9 Class 11 Sequences and Series
  2. Serial order wise
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Ex 9.3, 29 If A and G be A.M. and G.M., respectively between two positive numbers, prove that the numbers are A ± √((A+G)(A−G)) Let a & b be two numbers We need to show that the numbers are A ± √((A+G)(A−G)) i.e. a = A + √((𝐴+𝐺)(𝐴 −𝐺)) b = A – √((𝐴+𝐺)(𝐴 −𝐺)) Now we know that Arithmetic mean =A = ( a+b)/2 Geometric mean =G= √ab Putting value of A and G in RHS we can prove it is equal to a and b Solving A ± √((𝐴+𝐺)(𝐴 −𝐺)) = A ± √(𝐴2 −𝐺2) Putting A = (𝑎 + 𝑏)/2 & G = √𝑎𝑏 = ((𝑎 + 𝑏 )/2) ± √(((𝑎 + 𝑏 )/2)^2 −(√𝑎𝑏)2) = ((𝑎 + 𝑏 )/2) ± √(((𝑎 + 𝑏)2 )/4−𝑎𝑏) = ((𝑎 + 𝑏 )/2) ± √((𝑎2+𝑏2+2𝑎𝑏 −4𝑎𝑏)/4) = ((𝑎 + 𝑏 )/2) ± √((𝑎2 + 𝑏2 − 2𝑎𝑏 )/4) = (𝑎+𝑏)/2 ± √((𝑎 − 𝑏)2/4) = (𝑎+𝑏)/2 ± √(((𝑎 − 𝑏 )/2)^2 ) = (𝑎+𝑏)/2 ± (𝑎 − 𝑏 )/2 Thus, A + √((𝐴+𝐺)(𝐴 −𝐺)) = a & A – √((𝐴+𝐺)(𝐴 −𝐺)) = b Hence proved.

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