Example 15 - Chapter 9 Class 11 Sequences and Series
Last updated at May 29, 2018 by Teachoo
Transcript
Example 15 Find the sum of the sequence 7, 77, 777, 7777, ... to n terms. 7, 77, 777, 7777, ... n terms Here, 77/7 = 11 & 777/77 = 10.09 Thus, ( )/( ) ( )/( ) i.e. common ratio is not same This is not a GP We need to find sum Sum = 7 + 77 + 777 + 7777 + ...upto n terms = 7(1 + 11 + 111 + . upto n terms) = 7(1 + 11 + 111 + . upto n terms) Multiplying & dividing by 9 = 7/9 [9(1 + 11 + 111 + upto n term) = 7/9 [9 + 99 + 999 + 9999 + upto n terms] = 7/9 [(10 1) + (100 1) + (1000 1) + upto n terms] = 7/9 [(10 + 100 + 1000 + .n terms) 1 1 1 upto n terms] = 7/9 [(10 + 100 + 1000 + .n terms) (1 + 1 + 1 + upto n terms)] = 7/9 [(10 + 100 + 1000 + .n terms) n 1] = 7/9 [(10 + 100 + 1000 + .n terms) n] Now, a = 10, r = 10 For, r > 1 i.e. Sn = (a( ^ 1))/( 1) Putting value of a = 10 & r = 10 Sn = (10( 10 ^ 1))/(10 1) Sn = (10( 10 ^ 1))/9 Now substituting this value in (1) Sum = 7/9 [(10 + 102 + 103 + upto n terms) n] Sum = 7/9 [(10( 10 ^ 1))/9 " n" ] Thus, 7, 77, 777, 7777, ...upto n terms = 7/9 [(10( 10 ^ 1))/9 " n" ]
Examples
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Example 6 Important
Example 7
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Example 11
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Example 13
Example 14 Important
Example 15 Important You are here
Example 16
Example 17 Important
Example 18 Important
Example 19 Important
Example 20
Example 21 Important
Example 22
Example 23 Important
Example 24 Important
Chapter 9 Class 11 Sequences and Series
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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.