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Permutations And Combinations

Question
CBSEENMA11013102

Let S be the sum, P the product and R the sum of the reciprocals of n terms in a G.P. Prove that straight P squared straight R to the power of straight n space equals space straight S to the power of straight n.

Solution

Let a be the first term and r be the common ratio of a G.P.
                           S = the sum of n terms of G.P. = a + ar + .........arn-1

∴                          straight S equals fraction numerator straight a left parenthesis straight r to the power of straight n minus 1 right parenthesis over denominator straight r minus 1 end fraction                                                                  ...(i)
                           P = the product of n terms of G.P. = a. ar. ar2  .......  arn-1


∴                          P = straight a to the power of straight n straight r to the power of 1 plus 2 plus..... plus left parenthesis straight n minus 1 right parenthesis end exponent space equals space straight a to the power of straight n straight r to the power of fraction numerator straight n minus 1 over denominator 2 end fraction left parenthesis 1 plus straight n minus 1 right parenthesis end exponent
rightwards double arrow                       straight P space equals space straight a to the power of straight n straight r to the power of fraction numerator straight n left parenthesis straight n minus 1 right parenthesis over denominator 2 end fraction end exponent                                                                   ...(ii)
      R = the sum of the reciprocals of n terms of G.P. = space space 1 over straight a plus 1 over ar plus 1 over ar squared plus... space plus space 1 over ar to the power of straight n minus 1 end exponent
          = WiredFaculty
Now,          straight S over straight R space equals space fraction numerator begin display style fraction numerator straight a left parenthesis straight r to the power of straight n minus 1 right parenthesis over denominator straight r minus 1 end fraction end style over denominator begin display style fraction numerator straight r over denominator straight a left parenthesis straight r minus 1 right parenthesis space end fraction end style open parentheses begin display style fraction numerator straight r to the power of straight n minus 1 over denominator straight r to the power of straight n end fraction end style close parentheses end fraction space equals space fraction numerator straight a squared straight r to the power of straight n over denominator straight r end fraction space equals space straight a squared straight r to the power of straight n minus 1 end exponent
L.H.S.    = open parentheses straight S over straight R close parentheses to the power of straight n space equals space left parenthesis straight a squared straight r to the power of straight n minus 1 end exponent right parenthesis to the power of straight n space equals space straight a to the power of 2 straight n end exponent space straight r to the power of straight n left parenthesis straight n minus 1 right parenthesis end exponent
R.H.S. = straight P squared space equals space open square brackets straight a to the power of straight n space space straight r to the power of fraction numerator straight n left parenthesis straight n minus 1 right parenthesis over denominator 2 end fraction end exponent close square brackets squared space equals space straight a to the power of 2 straight n space space space end exponent straight r to the power of straight n left parenthesis straight n minus 1 right parenthesis end exponent

∴     L.H.S. = R.H.S.
Hence,   straight P squared straight R to the power of straight n space equals space straight S to the power of straight n.

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