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Ex 9.3, 15 - Given a GP with a = 729 and 7th term 64 - Ex 9.3

  1. Chapter 9 Class 11 Sequences and Series
  2. Serial order wise
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Ex9.3, 15 Given a G.P. with a = 729 and 7th term 64, determine S7. First term a = 729 and 7th term = 64 we know that nth term of G.P. = arn-1 a7 = ar6 putting values 64 = 729 r6 64/729 = r6 2^6/3^6 = r6 (2/3)^6= r6 (2/3)^6= r6 Comparing powers r = 2/3 We need to find sum of first 7 terms We know that Sum of n terms = 𝑎(1 − 𝑟^𝑛 )/(1 −𝑟) Sn = 𝑎(1 − 𝑟^𝑛 )/(1 −𝑟) S7 = 𝑎(1 −𝑟7)/(1 −𝑟) Putting values S7 = 729(1 − (2/3)^7 )/(1 − 2/3) S7 = 729(1 − (2/3)^7 )/(1 − 2/3) S7 = 729(1 − 128/2187)/((3 − 2)/3) S7 = 729(1 − 128/2187)/(1/3) S7 = 729(2059/2187) × 3 S7 = 2187(2059/2187) S7 = 2059 Hence S7 = 2059 Hence, sum of first 7 terms is 2059

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