Misc 19 - Ratio of AM and GM of a, b is m : n. Show that - AM and GM (Arithmetic Mean And Geometric mean)

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Misc 19 The ratio of the A.M and G.M. of two positive numbers a and b, is m: n. Show that a : b = (m + โˆš(๐‘š^2โˆ’๐‘›^2 )) : (m โ€“ โˆš(๐‘š^2โˆ’๐‘›^2 ) ) 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 Misc 19 The ratio of the A.M and G.M. of two positive numbers a and b, is m: n. Show that a : b = (m + โˆš(๐‘š^2โˆ’๐‘›^2 )) : (m โ€“ โˆš(๐‘š^2โˆ’๐‘›^2 ) ) Here, the two numbers be a and b. Arithmetic Mean =AM= (a+b)/2 & Geometric Mean=GM= โˆšab According to the question, AM/( GM" " ) = ๐‘š/๐‘› (๐‘Ž + ๐‘)/(2 โˆš๐‘Ž๐‘ " " ) = ๐‘š/๐‘› Applying componendo dividendo (๐‘Ž+๐‘+2โˆš๐‘Ž๐‘)/(๐‘Ž+๐‘ โˆ’2โˆš๐‘Ž๐‘) = (๐‘š + ๐‘›)/(๐‘š โˆ’ ๐‘› ) ((โˆš๐‘Ž)2+(โˆš๐‘)2+2(โˆš๐‘Žร—โˆš๐‘))/((โˆš๐‘Ž)2+(โˆš๐‘)2โˆ’2(โˆš๐‘Žร—โˆš๐‘) ) =(๐‘š + ๐‘›)/(๐‘š โˆ’ ๐‘› ) Using (x + y)2 = x2 + y2 + 2xy (x - y)2 = x2 + y2 - 2xy (โˆš๐‘Ž + โˆš๐‘)2/(โˆš๐‘Ž โˆ’ โˆš๐‘)2 = (๐‘š + ๐‘›)/(๐‘š โˆ’ ๐‘› ) ((โˆš๐‘Ž + โˆš๐‘)/(โˆš๐‘Ž โˆ’ โˆš๐‘))^2 = (๐‘š + ๐‘›)/(๐‘š โˆ’ ๐‘› ) (โˆš๐‘Ž + โˆš๐‘)/(โˆš๐‘Ž โˆ’ โˆš๐‘) = โˆš((๐‘š + ๐‘›)/(๐‘š โˆ’ ๐‘› )) (โˆš๐‘Ž + โˆš๐‘)/(โˆš๐‘Ž โˆ’ โˆš๐‘) = โˆš(๐‘š + ๐‘›)/(โˆš(๐‘š โˆ’ ๐‘›) ) Applying componendo dividendo ((โˆš๐‘Ž + โˆš๐‘) + (โˆš๐‘Ž โˆ’ โˆš๐‘))/((โˆš๐‘Ž + โˆš๐‘) โˆ’ (โˆš๐‘Ž โˆ’ โˆš๐‘) ) = (โˆš(๐‘š + ๐‘›) + โˆš(๐‘š โˆ’ ๐‘›))/(โˆš(๐‘š + ๐‘›) โˆ’โˆš(๐‘š โˆ’ ๐‘›)) (2โˆš๐‘Ž)/(2โˆš๐‘) = (โˆš(๐‘š + ๐‘›) + โˆš(๐‘š โˆ’ ๐‘›))/(โˆš(๐‘š + ๐‘›) โˆ’โˆš(๐‘š โˆ’ ๐‘›)) โˆš๐‘Ž/โˆš๐‘ = (โˆš(๐‘š + ๐‘›) + โˆš(๐‘š โˆ’ ๐‘›))/(โˆš(๐‘š + ๐‘›) โˆ’โˆš(๐‘š โˆ’ ๐‘›)) Squaring both sides (โˆš๐‘Ž/โˆš๐‘)^2 = ((โˆš(๐‘š + ๐‘›) + โˆš(๐‘š โˆ’ ๐‘›))/(โˆš(๐‘š + ๐‘›) โˆ’โˆš(๐‘š โˆ’ ๐‘›)))^2 (โˆš๐‘Ž)^2/(โˆš๐‘)^2 = (โˆš(๐‘š + ๐‘›) + โˆš(๐‘š โˆ’ ๐‘›))^2/(โˆš(๐‘š + ๐‘›) โˆ’โˆš(๐‘š โˆ’ ๐‘›))^2 Using (x + y)2 = x2 + y2 + 2xy (x - y)2 = x2 + y2 - 2xy ๐‘Ž/๐‘ = ((โˆš(๐‘š + ๐‘›) )^2+(โˆš(๐‘š โˆ’ ๐‘›) )^2+ 2(โˆš(๐‘š+๐‘›))(โˆš(๐‘šโˆ’๐‘›)))/((โˆš(๐‘š + ๐‘›) )^2+(โˆš(๐‘š โˆ’ ๐‘›) )^2โˆ’ 2(โˆš(๐‘š+๐‘›))(โˆš(๐‘šโˆ’๐‘›)) ) ๐‘Ž/๐‘ = (๐‘š + ๐‘› + ๐‘š โˆ’ ๐‘› + 2โˆš((๐‘š+๐‘›)(๐‘šโˆ’๐‘›) ))/(๐‘š + ๐‘› + ๐‘š โˆ’ ๐‘› โˆ’ 2โˆš((๐‘š+๐‘›)(๐‘šโˆ’๐‘›) )) ๐‘Ž/๐‘ = (๐‘š +๐‘š + ๐‘› โˆ’ ๐‘› + 2โˆš((๐‘š^2โˆ’ ๐‘›^2 ) ))/(๐‘š +๐‘š + ๐‘› โˆ’ ๐‘› โˆ’ 2โˆš((๐‘š^2โˆ’ ๐‘›^2 ) )) ๐‘Ž/๐‘ = (2๐‘š + 2โˆš((๐‘š^2โˆ’ ๐‘›^2 ) ))/(2๐‘š โˆ’ 2โˆš((๐‘š^2โˆ’ ๐‘›^2 ) )) ๐‘Ž/๐‘ = 2(๐‘š + โˆš((๐‘š^2โˆ’ ๐‘›^2 ) ))/2(๐‘š โˆ’ โˆš((๐‘š^2โˆ’ ๐‘›^2 ) )) ๐‘Ž/๐‘ = (๐‘š + โˆš((๐‘š^2โˆ’ ๐‘›^2 ) ))/(๐‘š โˆ’ โˆš((๐‘š^2โˆ’ ๐‘›^2 ) )) Thus, a : b = (m + โˆš(๐‘š^2โˆ’๐‘›^2 )) : (m โ€“ โˆš(๐‘š^2โˆ’๐‘›^2 ) ) Hence proved

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