Example 11 - Chapter 9 Class 11 Sequences and Series
Last updated at May 29, 2018 by Teachoo
Transcript
Example 11, In a G.P., the 3rd term is 24 and the 6th term is 192. Find the 10th term. 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, 3rd term is 24 i.e. a3 = 24 Putting an = 24, n = 3 in an formula 24 = a 𝑟^(3−1) 24 = ar2 ar2 = 24 Similarly, Given 6th term is 192 i.e. a6 = 192 Putting an = 192, n = 6 in an formula 192 = a 𝑟^(6−1) 192 = ar5 ar5 = 192 Now, our equations are ar2 = 24 …(1) & ar5 = 192 …(2) Dividing (2) by (1), ar5/ar2 = 192/24 𝑟5/𝑟2 = 8 𝑟5/𝑟2 = 8 r5 – 2 = 8 r3 = 8 r3 = 2 × 2 × 2 r3 = (2)3 r = 2. putting r in (1) ar2 = 24 a(2)2 = 24 a (4) = 24 a = 24/4 a = 6 Thus, a = 6 & r = 2 Now we need to find 10th term, we use the formula an = arn – 1 Putting n = 10 , a = 6, r = 2 a10 = ar10 – 1 = (6)(2)10 – 1 = 6 (2)9 = 3072 Hence, 10th term of GP is 3072
Examples
Example 1
Example 2
Example 3
Example 4
Example 5
Example 6 Important
Example 7
Example 8
Example 9
Example 10
Example 11 You are here
Example 12
Example 13
Example 14 Important
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
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.