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 You are here

Misc 22

Misc 23

Misc 24

Misc 25 Important

Misc 26

Misc 27

Misc 28 Important

Misc 29

Misc 30

Misc 31

Misc 32 Important

Chapter 9 Class 11 Sequences and Series

Serial order wise

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.