Last updated at May 29, 2018 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)^( 1) 128/2 = ( 2)^( 1) 64 = ( 2)^( 1) 26 = ( 2)^( 1) Squaring both sides (26)2 =[( 2)^( 1) ]^2 26 2 = [( 2)^2 ]^( 1) 212 = [2]^( 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 )^( 1) 729 = ( 3 )^(1+ 1) 729 = ( 3 )^ 3^6 = ( 3 )^ Squaring both sides (36)2 = ( 3)^2 36 2 = [( 3)^2 ]^ 312 = [3]^ 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)^( 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
Ex 9.3
Ex 9.3, 2
Ex 9.3, 3 Important
Ex 9.3, 4
Ex 9.3, 5 You are here
Ex 9.3, 6
Ex 9.3, 7
Ex 9.3, 8
Ex 9.3, 9
Ex 9.3, 10
Ex 9.3, 11 Important
Ex 9.3, 12
Ex 9.3, 13
Ex 9.3, 14
Ex 9.3, 15
Ex 9.3, 16
Ex 9.3, 17 Important
Ex 9.3, 18 Important
Ex 9.3, 19
Ex 9.3, 20
Ex 9.3, 21
Ex 9.3, 22 Important
Ex 9.3, 23
Ex 9.3, 24
Ex 9.3, 25
Ex 9.3, 26
Ex 9.3, 27 Important
Ex 9.3, 28 Important
Ex 9.3, 29 Important
Ex 9.3, 30
Ex 9.3, 31
Ex 9.3, 32
Ex 9.3
About the Author