Example 13 - Chapter 9 Class 11 Sequences and Series
Last updated at Dec. 8, 2016 by Teachoo
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
Examples
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Example 6 Important
Example 7
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Example 9
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Example 11
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Example 13 You are here
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Example 15 Important
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
Serial order wise
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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.