Solve all your doubts with Teachoo Black (new monthly pack available now!)
Are you in school? Do you love Teachoo?
We would love to talk to you! Please fill this form so that we can contact you
Ex 9.3
Ex 9.3, 2
Ex 9.3, 3 Important
Ex 9.3, 4
Ex 9.3, 5 (a)
Ex 9.3, 5 (b) Important
Ex 9.3, 5 (c)
Ex 9.3, 6
Ex 9.3, 7 Important
Ex 9.3, 8
Ex 9.3, 9 Important
Ex 9.3, 10
Ex 9.3, 11 Important
Ex 9.3, 12
Ex 9.3, 13 You are here
Ex 9.3, 14 Important
Ex 9.3, 15
Ex 9.3, 16 Important
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 Important
Ex 9.3, 24
Ex 9.3, 25
Ex 9.3, 26 Important
Ex 9.3, 27 Important
Ex 9.3, 28
Ex 9.3, 29 Important
Ex 9.3, 30 Important
Ex 9.3, 31
Ex 9.3, 32 Important
Ex 9.3
Last updated at May 29, 2018 by Teachoo
Ex 9.3, 13 How many terms of G.P. 3, 32, 33, … are needed to give the sum 120? First term = a = 3 Common difference r = 32/3 = 3 We know that Sum of n terms is = (𝑎(𝑟^𝑛− 1))/(𝑟 − 1) We need to find number of terms required to give sum 120 Let sum of n terms of this G.P. = 120 We need to find n Sn = (a(𝑟^𝑛− 1))/(r−1) 120 = (a(𝑟^𝑛− 1))/(r−1) Putting value of a & r 120 = 3(3n − 1)/(3 − 1) 120 = 3(3n − 1)/2 120 × 2 = 3(3n – 1) 240 = 3(3n – 1) 240/3 = 3n – 1 80 = 3n – 1 80 + 1 = 3n 81 = 3n 34 = 3n Comparing powers n = 4 Hence sum of first four terms is 120