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CBSE Class 11-science Answered

question is attached.
Asked by sudhanshubhushanroy | 08 Dec, 2017, 06:33: AM
answered-by-expert Expert Answer
begin mathsize 16px style Let space square root of 4 straight A to the power of blank end root the space two space numbers space be space straight a space and space straight b.
Let space straight A thin space and space straight G space be space the space arithmetic space and space geometrix space means space between space straight a space and space straight b.
Then comma
straight A equals fraction numerator straight a plus straight b over denominator 2 end fraction rightwards double arrow straight a plus straight b equals 2 straight A
And comma space straight G equals square root of ab rightwards double arrow straight G squared equals ab
The space equation space having space straight a space and space straight b space as space its space roots space is
straight x squared minus left parenthesis straight a plus straight b right parenthesis straight x plus ab equals 0
rightwards double arrow straight x squared minus 2 Ax plus straight G squared equals 0
rightwards double arrow straight x equals fraction numerator 2 straight A plus-or-minus square root of 4 straight A squared minus 4 straight G squared end root over denominator 2 end fraction rightwards double arrow straight x equals straight A plus-or-minus square root of straight A squared minus straight G squared end root
SO comma space the space two space numbers space are space straight a equals straight A plus square root of straight A squared minus straight G squared end root space and space straight b equals straight A minus square root of straight A squared minus straight G squared end root
Given comma
straight A colon straight G equals straight m colon straight n
rightwards double arrow straight A equals λm space and space straight G equals λn space for space some space straight lambda
Substituting space the space values space of space straight A space and space straight G space in space space straight a equals straight A plus square root of straight A squared minus straight G squared end root space and space straight b equals straight A minus square root of straight A squared minus straight G squared end root comma space we space have
straight a over straight b equals fraction numerator λm plus square root of straight lambda squared straight m squared minus straight lambda squared straight n squared end root over denominator λm minus square root of straight lambda squared straight m squared minus straight lambda squared straight n squared end root end fraction
rightwards double arrow straight a over straight b equals fraction numerator straight m plus square root of straight m squared minus straight n squared end root over denominator straight m minus square root of straight m squared minus straight n squared end root end fraction end style
Answered by Rashmi Khot | 08 Dec, 2017, 09:42: AM

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