Ex 8.2, 13 - Chapter 8 Class 11 Sequences and Series

Last updated at April 16, 2024 by

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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

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.