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Ex 8.2, 11 - Chapter 8 Class 11 Sequences and Series

Last updated at May 29, 2023 by

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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