# Ex 9.3, 5

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Ex9.3, 5 Which term of the following sequences: 2, 2√2, 4,…… is 128 2, 2√2, 4,…… 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 = (2√2)/2 = √2 nth term = 128 Putting values nth term = arn-1 128 = (2)(√2)^(𝑛−1) 128/2 = (√2)^(𝑛−1) 64 = (√2)^(𝑛−1) 26 = (√2)^(𝑛−1) Squaring both sides (26)2 =[(√2)^(𝑛−1) ]^2 26 × 2 = [(√2)^2 ]^(𝑛−1) 212 = [2]^(𝑛−1) Comparing powers 12 = n – 1 12 + 1 = n 13 = n n = 13 Hence, 13th term of GP is 128 Ex 9.3, 5 Which term of the following sequences: (b) √(3,) 3, 3√3,……is 729 ? √(3,) 3, 3 √3,…. 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 = √3 common ratio r = 3/√3 = √3 nth term = 729 Putting values an = arn – 1 729 = (√3 ) (√3 )^(𝑛−1) 729 = (√3 )^(1+ 𝑛 − 1) 729 = (√3 )^𝑛 3^6 = (√3 )^𝑛 Squaring both sides (36)2 = (√3)^2𝑛 36 × 2 = [(√3)^2 ]^𝑛 312 = [3]^𝑛 Comparing powers 12 = n n = 12 Hence 12th term of GP is 729 Ex9.3, 5 Which term of the following sequences: (c) 1/3, 1/9, 1/27, ….. Is 1/19683 ? 1/3, 1/9, 1/27, ….. 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 = 1/3 Common ratio r = (1/9)/(1/3) nth term = 1/19683 Putting values nth term = arn-1 1/19683 = (1/3) (1/3)^(𝑛−1) 1/19683 = (1/3)^(1+ 𝑛 −1) " " 1/19683 " = " (1/3)^𝑛 " " (1/3^9 )" = " (1/3)^𝑛 " " (1/3)^9 "= " (1/3)^𝑛 Comparing powers 9 = n n = 9 Hence 9th term of GP is 1/19683

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Chapter 9 Class 11 Sequences and Series

Serial order wise

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CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .