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Permutations And Combinations

Question
CBSEENMA11013142

If a, b, c are in G.P. and x, y are the arithmetic means of a, b and b, c respectively,

then prove that   straight a over straight x plus straight c over straight y equals 2 and space space 1 over straight x plus 1 over straight y equals 2 over straight b

Solution

Since a, b, c are in G.P.
∴       straight b over straight a space equals space straight c over straight b
rightwards double arrow      straight b squared space equals space ac                                                                                     ...(i)
Since x is the A.M. between a and b
∴                      straight x space equals space fraction numerator straight a plus straight b over denominator 2 end fraction                                                                    ...(ii)
Since y is the A.M. between b and c
∴                       "<pre                                                                   ...(iii)

       straight a over straight x plus straight c over straight y space equals space fraction numerator straight a over denominator begin display style fraction numerator straight a plus straight b over denominator 2 end fraction end style end fraction plus fraction numerator straight c over denominator begin display style fraction numerator straight b plus straight c over denominator 2 end fraction end style end fraction space equals space fraction numerator 2 straight a over denominator straight a plus straight b end fraction plus fraction numerator 2 straight c over denominator straight b plus straight c end fraction space equals space 2 open square brackets fraction numerator straight a left parenthesis straight b plus straight c right parenthesis plus straight c left parenthesis straight a plus straight b right parenthesis over denominator left parenthesis straight a plus straight b right parenthesis left parenthesis straight b plus straight c right parenthesis end fraction close square brackets

                            WiredFaculty      [By using (i)]
                            equals 2 open square brackets fraction numerator ab plus bc plus 2 ac over denominator ab plus bc plus 2 ac end fraction close square brackets space equals 2
∴   straight a over straight x plus straight c over straight y equals 2
Now,   1 over straight x plus 1 over straight y space equals space fraction numerator 1 over denominator begin display style fraction numerator straight a plus straight b over denominator 2 end fraction end style end fraction plus fraction numerator 1 over denominator begin display style fraction numerator straight b plus straight c over denominator 2 end fraction end style end fraction space equals space fraction numerator 2 over denominator straight a plus straight b end fraction plus fraction numerator 2 over denominator straight b plus straight c end fraction space equals space 2 open square brackets fraction numerator straight b plus straight c plus straight a plus straight b over denominator left parenthesis straight a plus straight b right parenthesis left parenthesis straight b plus straight c right parenthesis end fraction close square brackets space equals space 2 open square brackets fraction numerator straight c plus straight a plus 2 straight b over denominator ab plus ac plus straight b squared plus bc end fraction close square brackets
                      space space space equals 2 open square brackets fraction numerator straight c plus straight a plus 2 straight b over denominator ab plus bc plus straight b squared plus straight b squared end fraction close square brackets                                      [By using (i)]

                          WiredFaculty
∴  WiredFaculty

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