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Ex 9.3, 5 - Chapter 9 Class 11 Sequences and Series - Part 6

Ex 9.3, 5 - Chapter 9 Class 11 Sequences and Series - Part 7

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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)^( 1) 1/19683 = (1/3)^(1+ 1) " " 1/19683 " = " (1/3)^ " " (1/3^9 )" = " (1/3)^ " " (1/3)^9 "= " (1/3)^ Comparing powers 9 = n n = 9 Hence 9th term of GP is 1/19683

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.