Misc 21 - Chapter 9 Class 11 Sequences and Series - Part 5

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Misc 21 - Chapter 9 Class 11 Sequences and Series - Part 6

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Misc 21 - Chapter 9 Class 11 Sequences and Series - Part 7 Misc 21 - Chapter 9 Class 11 Sequences and Series - Part 8

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

Transcript

Misc 21 Find the sum of the following series upto n terms: (ii) .6 +.66 +. 666 +… .6 +.66 +. 666 +… to n terms Here, 0.66/0.6 = (66/100)/(6/10) = 1.1 & 0.666/0.66 = (666/1000)/(66/100) = 1.009 Thus, (π‘†π‘’π‘π‘œπ‘›π‘‘ π‘‘π‘’π‘Ÿπ‘š)/(πΉπ‘–π‘Ÿπ‘ π‘‘ π‘‘π‘’π‘Ÿπ‘š) β‰  (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 not same ∴ This is not a GP We need to find sum Sum = 0.6 + 0.66 + 0.666 + … to n terms = 6 [0.1 + 0.11 + 0.111 +… to n terms ] = 6 [0.1 + 0.11 + 0.111 +… to n terms ] Multiplying & dividing by 9 = 6/9 Γ— 9[0.1 + 0.11 + 0.111 +… to n terms ] = 2/3 [0.9 + 0.99 + 0.999 +… to n terms ] = 2/3 [ (9/10) + (99/100)+ (999/1000)+… to n terms ] = 2/3 [((10 βˆ’ 1)/10) + ((100 βˆ’ 1)/100)+ ((1000 βˆ’ 1)/1000)+ …to n terms ] = 2/3[(1 βˆ’ 1/10)+ (1 βˆ’ 1/100)+ (1 βˆ’ 1/1000) + … to n terms ] = 2/3 [(1 + 1 +…….to n terms) βˆ’ (1/10 " + " 1/100 " + " 1/1000 " + … to n terms" )] = 2/3 [n Γ— 1 – (1/10 " + " 1/100 " + " 1/1000 " + … to n terms" )] Now, a = 1/10, r = 1/10 For, r < 1 i.e. Sn = (a(1 βˆ’π‘Ÿ^𝑛))/(1 βˆ’ π‘Ÿ) Putting value of a = 1/10 & r = 1/10 = (1/10 (1 βˆ’(1/10)^𝑛 ))/(1 βˆ’ 1/10) = (1/10 (1 βˆ’(1/10)^𝑛 ))/(9/10) = 1/10[1 – (10)^(βˆ’π‘›)] Γ— 10/9 = 1/9[1 – (10)^(βˆ’π‘›)] Thus, (1/10 " + " 1/100 " + " 1/1000 " + … to n terms" ) = 1/9[1 – (10)^(βˆ’π‘›)] Now, Sum = 2/3 [n – (1/10 " + " 1/102 " + " 1/103 " + … to n terms" )] Substitute (1/10 " + " 1/100 " + " 1/1000 " + … to n terms" ) = 1/9[1 – (10)^(βˆ’π‘›)] in (1) = 2/3 [n – 1/9[1 – (10)^(βˆ’π‘›)] = 2/3n – 2/3 Γ—1/9[1 – (10)^(βˆ’π‘›)] = 2/3 n – 2/27[1 – (10)^(βˆ’π‘›)] Hence , 0.6 + 0.66 + 0.666 + … to n terms = 2/3 n – 2/27[1 – (10)^(βˆ’π‘›)]

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Davneet Singh
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.