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

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

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Ex9.3, 11 Introduction (2 + 3k) At k = 1, 2 + 31 At k = 2, 2 + 32 .. … …. At k = 11, 2 + 311 Ex9.3, 11 We calculate (31 + 32 + 33 + … + 311) separately In 31 + 32 + 33 + … + 311 32/31 = 3 & 33/32 = 3 Thus, (π‘†π‘’π‘π‘œπ‘›π‘‘ π‘‘π‘’π‘Ÿπ‘š)/(πΉπ‘–π‘Ÿπ‘ π‘‘ π‘‘π‘’π‘Ÿπ‘š) = (π‘‡β„Žπ‘–π‘Ÿπ‘‘ π‘‘π‘’π‘Ÿπ‘š)/(π‘†π‘’π‘π‘œπ‘›π‘‘ π‘‘π‘’π‘Ÿπ‘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 ∴ Sn = (π‘Ž(π‘Ÿ^𝑛 βˆ’ 1))/(π‘Ÿ βˆ’ 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 ∴ Sn = (π‘Ž(π‘Ÿ^𝑛 βˆ’ 1))/(π‘Ÿ βˆ’ 1) Putting values a = 3 , r = 3, n = 11 S11 = (3[311βˆ’1])/(3 βˆ’ 1) S11 =(3[311βˆ’1])/2 Hence 31 + 32 + … + 311 = (3[311βˆ’1])/2 From (1) Putting 31 + 32 + … + 311 = (3[311βˆ’1])/2 = 22 + (3(311 βˆ’ 1))/2 =22 + 3/2(311 βˆ’1) Therefore,

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.