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Example 10 - Chapter 9 Class 11 Sequences and Series
Last updated at May 29, 2018 by Teachoo
Transcript
Example 10 Which term of the G.P., 2,8,32, ... up to n terms is 131072? Given G.P., 2,8,32, ...upto n terms We know that an = arn – 1 where an = nth term of GP n is the number of terms a is the first term r is the common ratio Here, First term a = 2 Common ratio r = 8/2 = 4 & nth term = 131072, we need to find n Now an = 131072 arn-1 = 131072 2 × 4n – 1 = 131072 4n – 1 = 131072/2 Putting values an = arn-1 131072 = 2 × 4n – 1 131072/2 = 4n – 1 4n – 1 = 131072/2 4n – 1 = 65536 4n – 1 = (4)8 Comparing powers n – 1 = 8 n = 8 + 1 n = 9. Hence 131072 is the 9th term of the G.P.
Examples
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Example 5
Example 6 Important
Example 7
Example 8
Example 9
Example 10 You are here
Example 11
Example 12
Example 13
Example 14 Important
Example 15 Important
Example 16
Example 17 Important
Example 18 Important
Example 19 Important Not in Syllabus - CBSE Exams 2021
Example 20 Not in Syllabus - CBSE Exams 2021
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 9 years. He provides courses for Maths and Science at Teachoo.