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Ex 9.3, 18 - Find sum to n terms of sequence 8, 88, 888, 8888 - Geometric Progression(GP): Formulae based

Ex 9.3, 18 - Chapter 9 Class 11 Sequences and Series - Part 2
Ex 9.3, 18 - Chapter 9 Class 11 Sequences and Series - Part 3 Ex 9.3, 18 - Chapter 9 Class 11 Sequences and Series - Part 4


Transcript

Ex9.3, 18 Find the sum to n terms of the sequence, 8, 88, 888, 8888 8, 88, 888, 8888 to n term This is not a GP but we can relate it to a GP By writing as Sum = 8 + 88 + 888 + 8888 + upto n terms = 8(1) + 8(11) + 8(111) + upto n term Taking 8 common = 8(1 + 11 + 111 + upto n term) Divide & multiply by 9 = 8/9[9(1 + 11 + 111 + upto n term)] = 8/9 [ 9 + 99 + 999 + 9999 + upto n terms] Sum = 8/9 [ 9 + 99 + 999 + 9999 + upto n terms] = 8/9 [ (10 1)+(100 1)+(1000 1)+ upto n terms] = 8/9 [ (10 1)+(102 1)+(103 1)+ upto n terms] = 8/9 [ (10 + 102 + 103 upto n terms) (1 + 1 + 1 + upto n terms)] = 8/9 [(10 + 102 + 103 upto n terms) n 1] We will solve (10 + 102 + 103 upto n terms) separately We can observe that this is GP With first term a = 10 & common ratio r = 102/10 = 10 We know that sum of n terms = (a( ^ 1))/( 1) i.e. Sn =(a( ^ 1))/( 1) putting value of a & r Sn = (10(10n 1))/(10 1) Substituting 10 + 102 + 103 + upto n times = (10(10n 1))/9 in (1) Sum = 8/9 [(10 + 102 + 103 + upto n terms) n] = 8/9 [ (10(10n 1))/(10 1) n ] = 8/9 [ (10(10n 1))/9 n ] = 8/9 [ (10(10n 1) 9n)/9] = 8/9 [ (10(10n 1) 9n)/9] = 8/9 [ 10(10n 1)/9 9/9n] = 8/9 10(10n 1)/9 8/9 9/9n] = 80/81(10n 1) - 8/9n Hence sum of sequence 8 + 88 + 888 + 8888 + .. to n terms = = 80/81(10n 1) 8/9n

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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.