Let S be the sum, P the product and R the sum of the reciprocals

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 Multiple Choice QuestionsShort Answer Type

81.

How many terms of the sequence

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82.

Sum the following series:

7 + 77 + 777 + ......... to n terms

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83.

Sum the following series:

0.5 + 0.55 + 0.555.......... to n terms



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84.

Given a G.P. with a = 729 and 7th term 64, determine S7.

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85.

Determine the number of terms in a GP., if t1 = 3, tn = 96 and Sn = 189.

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86.

Find the least value of n for which the sum 1 + 3 + 32+ ...... to n terms is greater than 7000.

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87.

The ratio of the sum of first three terms is to that of first 6 terms of a G.P. is 125 : 152. Find the common ratio.

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88.

If S1, S2 and S3 be respectively the sums of n, 2n and 3n terms of a GP., prove that S1 (S3 – S2) = (S2 – S1)2

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 Multiple Choice QuestionsLong Answer Type

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89.

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.


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

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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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 Multiple Choice QuestionsShort Answer Type

90.

Evaluate:

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