1. Chapter 9 Class 11 Sequences and Series (Term 1)
  2. Serial order wise
  3. Ex 9.3

Ex 9.3, 11 - Chapter 9 Class 11 Sequences and Series (Term 1)

Last updated at May 29, 2018 by

Ex 9.3, 11 - Evaluate sigma 1 to 11, 2 + 3k - Chapter 9 - Geometric Progression(GP): Formulae based

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Ex 9.3, 11 - Chapter 9 Class 11 Sequences and Series - Part 2

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Ex 9.3, 11 - Chapter 9 Class 11 Sequences and Series - Part 3 Ex 9.3, 11 - Chapter 9 Class 11 Sequences and Series - Part 4 Ex 9.3, 11 - Chapter 9 Class 11 Sequences and Series - Part 5

  1. Chapter 9 Class 11 Sequences and Series (Term 1)
  2. Serial order wise

    3. Ex 9.3

    Ex 9.4 →

Transcript

Ex9.3, 11 Introduction (2 + 3k) At k = 1, 2 + 31 At k = 2, 2 + 32 .. โ€ฆ โ€ฆ. At k = 11, 2 + 311 Ex9.3, 11 We calculate (31 + 32 + 33 + โ€ฆ + 311) separately In 31 + 32 + 33 + โ€ฆ + 311 32/31 = 3 & 33/32 = 3 Thus, (๐‘†๐‘’๐‘๐‘œ๐‘›๐‘‘ ๐‘ก๐‘’๐‘Ÿ๐‘š)/(๐น๐‘–๐‘Ÿ๐‘ ๐‘ก ๐‘ก๐‘’๐‘Ÿ๐‘š) = (๐‘‡โ„Ž๐‘–๐‘Ÿ๐‘‘ ๐‘ก๐‘’๐‘Ÿ๐‘š)/(๐‘†๐‘’๐‘๐‘œ๐‘›๐‘‘ ๐‘ก๐‘’๐‘Ÿ๐‘š) i.e. common ratio is same Thus, it is a G.P. First term = a = 3 Common ratio = 3^2/3^1 = 3 Since, r > 1 โˆด Sn = (๐‘Ž(๐‘Ÿ^๐‘› โˆ’ 1))/(๐‘Ÿ โˆ’ 1) where Sn = sum of n terms of GP n is the number of terms a is the first term & r is the common ratio โˆด Sn = (๐‘Ž(๐‘Ÿ^๐‘› โˆ’ 1))/(๐‘Ÿ โˆ’ 1) Putting values a = 3 , r = 3, n = 11 S11 = (3[311โˆ’1])/(3 โˆ’ 1) S11 =(3[311โˆ’1])/2 Hence 31 + 32 + โ€ฆ + 311 = (3[311โˆ’1])/2 From (1) Putting 31 + 32 + โ€ฆ + 311 = (3[311โˆ’1])/2 = 22 + (3(311 โˆ’ 1))/2 =22 + 3/2(311 โˆ’1) Therefore,

Chapter 9 Class 11 Sequences and Series (Term 1)
Serial order wise

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.