
Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Ex 8.2
Ex 8.2, 2
Ex 8.2, 3 Important
Ex 8.2, 4
Ex 8.2, 5 (a)
Ex 8.2, 5 (b) Important
Ex 8.2, 5 (c)
Ex 8.2, 6
Ex 8.2, 7 Important
Ex 8.2, 8
Ex 8.2, 9 Important
Ex 8.2, 10
Ex 8.2, 11 Important
Ex 8.2, 12
Ex 8.2, 13 You are here
Ex 8.2, 14 Important
Ex 8.2, 15
Ex 8.2, 16 Important
Ex 8.2, 17 Important
Ex 8.2, 18 Important
Ex 8.2, 19
Ex 8.2, 20
Ex 8.2, 21
Ex 8.2, 22 Important
Ex 8.2, 23 Important
Ex 8.2, 24
Ex 8.2, 25
Ex 8.2, 26 Important
Ex 8.2, 27 Important
Ex 8.2, 28
Ex 8.2, 29 Important
Ex 8.2, 30 Important
Ex 8.2, 31
Ex 8.2, 32 Important
Last updated at May 29, 2023 by Teachoo
Ex 8.2, 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