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Example 13 - Chapter 9 Class 11 Sequences and Series (Term 1)
Last updated at March 30, 2023 by Teachoo
Examples
Example 1 (i)
Example 1 (ii)
Example 2
Example 3 Important
Example 4 Deleted for CBSE Board 2023 Exams
Example 5 Deleted for CBSE Board 2023 Exams
Example 6 Important Deleted for CBSE Board 2023 Exams
Example 7 Deleted for CBSE Board 2023 Exams
Example 8 Important
Example 9
Example 10 Important
Example 11
Example 12 Important
Example 13 You are here
Example 14 Important
Example 15 Important
Example 16
Example 17 Important
Example 18 Important
Example 19 Important Deleted for CBSE Board 2023 Exams
Example 20 Deleted for CBSE Board 2023 Exams
Example 21 Important
Example 22
Example 23
Example 24 Important
Chapter 9 Class 11 Sequences and Series
Serial order wise
- Ex 9.1
- Ex 9.2
- Ex 9.3
- Ex 9.4
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Examples
- Miscellaneous
Transcript
Example 13, How many terms of the G.P. 3, 3/2, 3/4 , , ,... are needed to give the sum 3069/512 ? Here First term = a = 3, Common ratio r = (3/2)/3 = 3/(2 3) = 1/2 We know that sum of n term = ( ( 1 ^ ))/(1 ) Sn = (a(1 ^ ))/(1 r) Given that Sn = 3069/512 & we need to find n. 3069/512 = (a(1 ^ ))/(1 ) 3069/512 = (3[1 (1/2)^ ])/(1 1/2) 3069/512 =(3[1 (1/2)^ ])/( 1/2) 3069/512 = 6[1 (1/2)^ ] 3069/(512 6) = 1 (1/2)^ 3069/3072 = 1 (1/2)^ 1 (1/2)^ = 3069/3072 " " (1/2)^ " = " ("1 " 3069/3072) (1/2)^ = ((3072 3069)/3072) (1/2)^ = (3/3072) (1/2)^ = (1/1024) (1/2)^ = (1/2)10 Comparing powers n = 10 Hence 10 terms are needed to give sum 3069/512
