Ex 9.3, 28 - Sum of two numbers is 6 times their geometric mean - Geometric Mean (GM)

  1. Chapter 9 Class 11 Sequences and Series
  2. Serial order wise
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Ex9.3, 28 The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio (3 + 2 √2) :"(3 − 2 " √2) Introduction Componendo dividendo If 𝑥/𝑦 = 𝑎/𝑏 Applying componendo dividendo (𝑥 + 𝑦)/(𝑥 − 𝑦) = (𝑎 + 𝑏)/(𝑎 − 𝑏) Eg: Taking 1/2 = 4/8 (1+ 2)/(1 − 2) = (4 + 8)/(4 − 8) 3/(−1) = 12/(−4) -3 = -3 Ex 9.3, 28 The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio (3 + 2 √2) :"(3 − 2 " √2) Let a & b be the numbers We know that Geometric mean of two numbers a & b is √𝑎𝑏 i.e. GM of a & b = √𝑎𝑏 According to the question Sum of two numbers a and b is 6 times of their GM a + b = 6√𝑎𝑏 Solving, (𝑎 + 𝑏 )/(2√𝑎𝑏) = 3/1 Applying componendo & dividendo (𝑎+𝑏+2√𝑎𝑏)/(𝑎+𝑏 −2√𝑎𝑏) = (3 + 1)/(3 − 1 ) ((√𝑎)2+(√𝑏)2+2(√𝑎×√𝑏))/((√𝑎)2+(√𝑏)2−2(√𝑎×√𝑏) ) = 4/2 Using (x + y)2 = x2 + y2 + 2xy (x - y)2 = x2 + y2 - 2xy (√𝑎 + √𝑏)2/(√𝑎 − √𝑏)2 = 2/1 ((√𝑎 + √𝑏)/(√𝑎 − √𝑏))^2 = 2/1 (√𝑎 + √𝑏)/(√𝑎 − √𝑏) = √2/( 1) Again applying componendo & dividendo ((√𝑎+ √𝑏)+(√𝑎 − √𝑏))/((√𝑎 + √𝑏) −(√𝑎 − √𝑏) ) = (√2 + 1)/(√2 − 1) (√𝑎 +√𝑎 + √𝑏 − √𝑏)/(√𝑎 + √𝑏 − √𝑎 + √𝑏) = (√2 + 1)/(√2 − 1) (2√𝑎 + 0)/(√𝑏 + √𝑏 + √𝑎 − √𝑎) = (√2 + 1)/(√2 − 1) (2√𝑎)/(2√𝑏 + 0) = (√2 + 1)/(√2 − 1) (2√𝑎)/(2√𝑏 ) = (√2 + 1)/(√2 − 1) √(𝑎/𝑏) = (√2 + 1)/(√2 − 1) Squaring both sides (√(𝑎/𝑏))^2 = ((√2 + 1)/(√2 − 1))^2 𝑎/𝑏 = ((√2 + 1)2)/((√2 − 1)2) 𝑎/𝑏 = ((√2)2 + (1)2 + 2√2 × 1)/((√2)2 + (1)2 − 2√2 × 1) 𝑎/𝑏 = (2 + 1 + 2√2)/(2 + 1 − 2√2) 𝑎/𝑏 = (3 + 2√2)/(3 − 2√2) Thus the ratio of a & b is 3 + 2√3: 3 – 2√2 Hence proved

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