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Misc 21 Find sum (i) 5 + 55 + 555 (ii) 0.6 + 0.66 + 0.666 - Miscellaneous

Misc 21 - Chapter 9 Class 11 Sequences and Series - Part 2
Misc 21 - Chapter 9 Class 11 Sequences and Series - Part 3
Misc 21 - Chapter 9 Class 11 Sequences and Series - Part 4

 

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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/9

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.