Example 10 - Chapter 9 Class 11 Sequences and Series (Term 1)
Last updated at Sept. 3, 2021 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 You are here
Example 11
Example 12 Important
Example 13
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 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.
