Ex 5.3, 3 (ix) - Chapter 5 Class 10 Arithmetic Progressions

Last updated at April 16, 2024 by

Ex 5.3, 3 - Chapter 5 Class 10 Arithmetic Progressions - Part 23

Ex 5.3, 3 - Chapter 5 Class 10 Arithmetic Progressions - Part 24

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Ex 5.3, 3 In an AP (ix) Given a = 3, n = 8, S = 192, find d. Given a = 3, n = 8 , Sn = 192 We know that Sn = 𝑛/2 (2𝑎+(𝑛−1)𝑑) Putting a = 3, n = 8 , Sn = 192 192 = 8/2(2×3+(8−1)×𝑑) 192 = 4 (6 + 7 d) 192/4 = 6 + 7d 48 = 6 + 7d 48 – 6 = 7d 42 = 7d 42/7=𝑑 6 = d d = 6

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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 and Computer Science at Teachoo