a, b, c are in continued  proportion , prove that  a\c =   2a2−5ab+7b2\2b2−5bc+ 7c2

Asked by tafeelsayyed786 | 19th Sep, 2020, 01:56: PM

Expert Answer:

Given: a, b and c are in continued proportion
straight a over straight b equals straight b over straight c rightwards double arrow straight b squared equals ac
Consider comma
fraction numerator 2 straight a squared minus 5 ab plus 7 straight b squared over denominator 2 straight b squared minus 5 bc plus 7 straight c squared end fraction equals fraction numerator 2 straight a squared minus 5 ab plus 7 ac over denominator 2 ac minus 5 bc plus 7 straight c squared end fraction equals fraction numerator straight a open parentheses 2 straight a minus 5 straight b plus 7 straight c close parentheses over denominator straight c open parentheses 2 straight a minus 5 straight b plus 7 straight c close parentheses end fraction equals straight a over straight c
Hence comma space straight a over straight c equals fraction numerator 2 straight a squared minus 5 ab plus 7 straight b squared over denominator 2 straight b squared minus 5 bc plus 7 straight c squared end fraction

Answered by Renu Varma | 22nd Sep, 2020, 02:11: PM