1. Chapter 8 Class 11 Sequences and Series
  2. Serial order wise
  3. Miscellaneous
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Misc 2 - Chapter 8 Class 11 Sequences and Series

Last updated at Dec. 16, 2024 by Teachoo

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Misc 2 The sum of some terms of G.P. is 315 whose first term and the common ratio are 5 and 2, respectively. Find the last term and the number of terms. Let a be the first term of GP & r be the common ratio of GP It is given that a = 5 & r = 2 (r > 1) Sum of some term of a GP = 315 Let the sum of n terms of GP = 315 Sn = 315 We know that Sum of n terms of GP = (a( ^ 1))/(r 1) Sn = (a( ^ 1))/(r 1) 315 = (a( ^ 1))/(r 1) Putting r = 2 & a = 5 315 = (5(2n 1))/(2 1) 315 = (5(2n 1))/1 315 = 5 (2n 1) 315/5 = 2n 1 63 = 2n 1 64 = 2n (2)6 = 2n comparing powers n = 6 Hence number of terms is 6 We need to find last term of GP Let l be the last term of GP We know that n term of GP an = arn 1 l = arn 1 Putting a = 5 , r = 2 & n = 6 l = 5(2)6 1 = 5(2)5 = 5(2 2 2 2 2) = 5 64 = 320 Hence, the last term is 320
  1. Chapter 8 Class 11 Sequences and Series
  2. Serial order wise

    3. Miscellaneous

    Arithmetic Progression →
Arithmetic Progression →
Chapter 8 Class 11 Sequences and Series
Serial order wise

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo