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Last updated at Dec. 8, 2016 by Teachoo

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Misc 21 Find the sum of the following series upto n terms: (i) 5 + 55 + 555 + … This is not GP but it can relate it to a GP by writing as Sum = 5 + 55 + 555 + ….. to n terms = 5(1) + 5(11) + 5(111) + … to n terms taking 5 common = 5(1 + 11 + 111 + … to n term) Divide & multiply by 9 = 5/9[9(1 + 11 + 111 + … to n term)] = 5/9 [9 + 99 + 999 +….to n terms ] = 5/9 [(10 – 1) + (100 – 1)+ (1000 – 1) + … to n terms ] = 5/9 [(10 – 1) + (102 – 1)+ (103 – 1) + … to n terms ] = 5/9 [(10 – 1) + (102 – 1)+ (103 – 1) + … to n terms ] = 5/9 [(10 + 102 + 103 + …. n terms) – (1 + 1 + 1 + … n terms) = 5/9 [(10 + 102 + 103 + …. n terms) – n × 1] = 5/9 [(10 + 102 + 103 + …. n terms) – n] We will solve (10 + 102 + 103 + …. n terms) separately We can observe that this is GP where first term = a = 10 & common ratio r = 102/10 = 10 We know that sum of n terms = (a(𝑟^𝑛− 1))/(𝑟 − 1) Putting value of a & r 10 + 102 + 103 + …….to n terms = (10(10n − 1))/(10 − 1) Substitute 10 + 102 + 103 + …….to n terms = (10(10n − 1))/(10 − 1) in (1) Sum = 5/9 [(10 + 102 + 103 + …….to n terms ) – n] = 5/9 [ (10(10n−1))/(10−1) − n ] = 5/9 [ (10(10n−1))/9 − n ] = 50/81 (10n – 1) − 5n/9 Hence sum of sequence 5 + 55 + 555 + ….. to n terms is = 50/81 (10n – 1) - 5n/9Misc 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, (𝑆𝑒𝑐𝑜𝑛𝑑 𝑡𝑒𝑟𝑚)/(𝐹𝑖𝑟𝑠𝑡 𝑡𝑒𝑟𝑚) ≠ (𝑇ℎ𝑖𝑟𝑑 𝑡𝑒𝑟𝑚)/(𝑆𝑒𝑐𝑜𝑛𝑑 𝑡𝑒𝑟𝑚) 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 × 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 − 1)/10) + ((100 − 1)/100)+ ((1000 − 1)/1000)+ …to n terms ] = 2/3[(1 − 1/10)+ (1 − 1/100)+ (1 − 1/1000) + … to n terms ] = 2/3 [(1 + 1 +…….to n terms) − (1/10 " + " 1/100 " + " 1/1000 " + … to n terms" )] = 2/3 [n × 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 −𝑟^𝑛))/(1 − 𝑟) Putting value of a = 1/10 & r = 1/10 = (1/10 (1 −(1/10)^𝑛 ))/(1 − 1/10) = (1/10 (1 −(1/10)^𝑛 ))/(9/10) = 1/10[1 – (10)^(−𝑛)] × 10/9 = 1/9[1 – (10)^(−𝑛)] Thus, (1/10 " + " 1/100 " + " 1/1000 " + … to n terms" ) = 1/9[1 – (10)^(−𝑛)] 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)^(−𝑛)] in (1) = 2/3 [n – 1/9[1 – (10)^(−𝑛)] = 2/3n – 2/3 ×1/9[1 – (10)^(−𝑛)] = 2/3 n – 2/27[1 – (10)^(−𝑛)] Hence , 0.6 + 0.66 + 0.666 + … to n terms = 2/3 n – 2/27[1 – (10)^(−𝑛)]

Miscellaneous

Misc 1

Misc 2

Misc 3 Important

Misc 4

Misc 5

Misc 6

Misc 7 Important

Misc 8

Misc 9

Misc 10

Misc 11

Misc 12

Misc 13

Misc 14

Misc 15

Misc 16 Important

Misc 17

Misc 18

Misc 19 Important

Misc 20

Misc 21 Not in Syllabus - CBSE Exams 2021 You are here

Misc 22

Misc 23 Not in Syllabus - CBSE Exams 2021

Misc 24 Not in Syllabus - CBSE Exams 2021

Misc 25 Important Not in Syllabus - CBSE Exams 2021

Misc 26

Misc 27

Misc 28 Important

Misc 29

Misc 30

Misc 31

Misc 32 Important

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About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.