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Last updated at Aug. 28, 2021 by Teachoo

Transcript

Misc 21 Find the sum of the following series upto n terms: (ii) .6 +.66 +. 666 +β¦ .6 +.66 +. 666 +β¦ 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, (ππππππ π‘πππ)/(πΉπππ π‘ π‘πππ) β (πβπππ π‘πππ)/(ππππππ π‘πππ) i.e. common ratio is not same β΄ This is not a GP We need to find sum Sum = 0.6 + 0.66 + 0.666 + β¦ to n terms = 6 [0.1 + 0.11 + 0.111 +β¦ to n terms ] = 6 [0.1 + 0.11 + 0.111 +β¦ to n terms ] Multiplying & dividing by 9 = 6/9 Γ 9[0.1 + 0.11 + 0.111 +β¦ to n terms ] = 2/3 [0.9 + 0.99 + 0.999 +β¦ to n terms ] = 2/3 [ (9/10) + (99/100)+ (999/1000)+β¦ to n terms ] = 2/3 [((10 β 1)/10) + ((100 β 1)/100)+ ((1000 β 1)/1000)+ β¦to n terms ] = 2/3[(1 β 1/10)+ (1 β 1/100)+ (1 β 1/1000) + β¦ to n terms ] = 2/3 [(1 + 1 +β¦β¦.to n terms) β (1/10 " + " 1/100 " + " 1/1000 " + β¦ to n terms" )] = 2/3 [n Γ 1 β (1/10 " + " 1/100 " + " 1/1000 " + β¦ to n terms" )] Now, a = 1/10, r = 1/10 For, r < 1 i.e. Sn = (a(1 βπ^π))/(1 β π) Putting value of a = 1/10 & r = 1/10 = (1/10 (1 β(1/10)^π ))/(1 β 1/10) = (1/10 (1 β(1/10)^π ))/(9/10) = 1/10[1 β (10)^(βπ)] Γ 10/9 = 1/9[1 β (10)^(βπ)] Thus, (1/10 " + " 1/100 " + " 1/1000 " + β¦ to n terms" ) = 1/9[1 β (10)^(βπ)] Now, Sum = 2/3 [n β (1/10 " + " 1/102 " + " 1/103 " + β¦ to n terms" )] Substitute (1/10 " + " 1/100 " + " 1/1000 " + β¦ to n terms" ) = 1/9[1 β (10)^(βπ)] in (1) = 2/3 [n β 1/9[1 β (10)^(βπ)] = 2/3n β 2/3 Γ1/9[1 β (10)^(βπ)] = 2/3 n β 2/27[1 β (10)^(βπ)] Hence , 0.6 + 0.66 + 0.666 + β¦ to n terms = 2/3 n β 2/27[1 β (10)^(βπ)]

Miscellaneous

Misc 1

Misc 2

Misc 3 Important

Misc 4

Misc 5

Misc 6 Important

Misc 7 Important

Misc 8

Misc 9

Misc 10 Important

Misc 11

Misc 12

Misc 13

Misc 14 Important

Misc 15

Misc 16 Important

Misc 17

Misc 18

Misc 19 Important

Misc 20

Misc 21 (i) Important

Misc 21 (ii) You are here

Misc 22 Important

Misc 23 Important

Misc 24 Deleted for CBSE Board 2022 Exams

Misc 25 Important Deleted for CBSE Board 2022 Exams

Misc 26 Deleted for CBSE Board 2022 Exams

Misc 27

Misc 28 Important

Misc 29 Important

Misc 30

Misc 31 Important

Misc 32 Important

About the Author

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.