Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Chapter 9 Class 11 Sequences and Series
Question 5 Important Deleted for CBSE Board 2024 Exams
Question 9 Important Deleted for CBSE Board 2024 Exams
Question 15 Important Deleted for CBSE Board 2024 Exams
Question 17 Deleted for CBSE Board 2024 Exams
Example 9 Important
Example 10 Important
Ex 8.2, 3 Important
Ex 8.2, 11 Important
Ex 8.2, 17 Important
Ex 8.2, 18 Important You are here
Ex 8.2, 22 Important
Ex 8.2, 28
Ex 8.2, 29 Important
Ex 9.4.4 Important Deleted for CBSE Board 2024 Exams
Question 7 Important Deleted for CBSE Board 2024 Exams
Question 9 Important Deleted for CBSE Board 2024 Exams
Question 10 Deleted for CBSE Board 2024 Exams
Question 9 Deleted for CBSE Board 2024 Exams
Question 9 Important Deleted for CBSE Board 2024 Exams
Misc 10 Important
Question 13 Important Deleted for CBSE Board 2024 Exams
Misc 14 Important
Misc 18 Important
Chapter 9 Class 11 Sequences and Series
Last updated at March 27, 2024 by Teachoo
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