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 โˆš2) :"(3 โˆ’ 2 " โˆš2) Introduction Componendo dividendo If ๐‘ฅ/๐‘ฆ = ๐‘Ž/๐‘ Applying componendo dividendo (๐‘ฅ + ๐‘ฆ)/(๐‘ฅ โˆ’ ๐‘ฆ) = (๐‘Ž + ๐‘)/(๐‘Ž โˆ’ ๐‘) Eg: Taking 1/2 = 4/8 (1+ 2)/(1 โˆ’ 2) = (4 + 8)/(4 โˆ’ 8) 3/(โˆ’1) = 12/(โˆ’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 โˆš2) :"(3 โˆ’ 2 " โˆš2) Let a & b be the numbers We know that Geometric mean of two numbers a & b is โˆš๐‘Ž๐‘ i.e. GM of a & b = โˆš๐‘Ž๐‘ According to the question Sum of two numbers a and b is 6 times of their GM a + b = 6โˆš๐‘Ž๐‘ Solving, (๐‘Ž + ๐‘ )/(2โˆš๐‘Ž๐‘) = 3/1 Applying componendo & dividendo (๐‘Ž+๐‘+2โˆš๐‘Ž๐‘)/(๐‘Ž+๐‘ โˆ’2โˆš๐‘Ž๐‘) = (3 + 1)/(3 โˆ’ 1 ) ((โˆš๐‘Ž)2+(โˆš๐‘)2+2(โˆš๐‘Žร—โˆš๐‘))/((โˆš๐‘Ž)2+(โˆš๐‘)2โˆ’2(โˆš๐‘Žร—โˆš๐‘) ) = 4/2 Using (x + y)2 = x2 + y2 + 2xy (x - y)2 = x2 + y2 - 2xy (โˆš๐‘Ž + โˆš๐‘)2/(โˆš๐‘Ž โˆ’ โˆš๐‘)2 = 2/1 ((โˆš๐‘Ž + โˆš๐‘)/(โˆš๐‘Ž โˆ’ โˆš๐‘))^2 = 2/1 (โˆš๐‘Ž + โˆš๐‘)/(โˆš๐‘Ž โˆ’ โˆš๐‘) = โˆš2/( 1) Again applying componendo & dividendo ((โˆš๐‘Ž+ โˆš๐‘)+(โˆš๐‘Ž โˆ’ โˆš๐‘))/((โˆš๐‘Ž + โˆš๐‘) โˆ’(โˆš๐‘Ž โˆ’ โˆš๐‘) ) = (โˆš2 + 1)/(โˆš2 โˆ’ 1) (โˆš๐‘Ž +โˆš๐‘Ž + โˆš๐‘ โˆ’ โˆš๐‘)/(โˆš๐‘Ž + โˆš๐‘ โˆ’ โˆš๐‘Ž + โˆš๐‘) = (โˆš2 + 1)/(โˆš2 โˆ’ 1) (2โˆš๐‘Ž + 0)/(โˆš๐‘ + โˆš๐‘ + โˆš๐‘Ž โˆ’ โˆš๐‘Ž) = (โˆš2 + 1)/(โˆš2 โˆ’ 1) (2โˆš๐‘Ž)/(2โˆš๐‘ + 0) = (โˆš2 + 1)/(โˆš2 โˆ’ 1) (2โˆš๐‘Ž)/(2โˆš๐‘ ) = (โˆš2 + 1)/(โˆš2 โˆ’ 1) โˆš(๐‘Ž/๐‘) = (โˆš2 + 1)/(โˆš2 โˆ’ 1) Squaring both sides (โˆš(๐‘Ž/๐‘))^2 = ((โˆš2 + 1)/(โˆš2 โˆ’ 1))^2 ๐‘Ž/๐‘ = ((โˆš2 + 1)2)/((โˆš2 โˆ’ 1)2) ๐‘Ž/๐‘ = ((โˆš2)2 + (1)2 + 2โˆš2 ร— 1)/((โˆš2)2 + (1)2 โˆ’ 2โˆš2 ร— 1) ๐‘Ž/๐‘ = (2 + 1 + 2โˆš2)/(2 + 1 โˆ’ 2โˆš2) ๐‘Ž/๐‘ = (3 + 2โˆš2)/(3 โˆ’ 2โˆš2) Thus the ratio of a & b is 3 + 2โˆš3: 3 โ€“ 2โˆš2 Hence proved

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.