Check sibling questions

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

Misc 21 - Chapter 9 Class 11 Sequences and Series - Part 6
Misc 21 - Chapter 9 Class 11 Sequences and Series - Part 7 Misc 21 - Chapter 9 Class 11 Sequences and Series - Part 8

This video is only available for Teachoo black users

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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‘›)]

Ask a doubt
Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.