Ex 5.3, 3 (ix) - Chapter 5 Class 10 Arithmetic Progressions (Term 2)
Last updated at Aug. 3, 2021 by
Transcript
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 = π/2 (2π+(πβ1)π) Putting a = 3, n = 8 , Sn = 192 192 = 8/2(2Γ3+(8β1)Γπ) 192 = 4 (6 + 7 d) 192/4 = 6 + 7d 48 = 6 + 7d 48 β 6 = 7d 42 = 7d 42/7=π 6 = d d = 6
Ex 5.3
Ex 5.3, 1 (i)
Ex 5.3, 1 (ii)
Ex 5.3, 1 (iii) Important
Ex 5.3, 1 (iv)
Ex 5.3, 2 (i)
Ex 5.3, 2 (ii)
Ex 5.3, 2 (iii) Important
Ex 5.3, 3 (i)
Ex 5.3, 3 (ii)
Ex 5.3, 3 (iii)
Ex 5.3, 3 (iv) Important
Ex 5.3, 3 (v)
Ex 5.3, 3 (vi) Important
Ex 5.3, 3 (vii)
Ex 5.3, 3 (viii) Important
Ex 5.3, 3 (ix) You are here
Ex 5.3, 3 (x)
Ex 5.3, 4
Ex 5.3, 5
Ex 5.3, 6 Important
Ex 5.3, 7
Ex 5.3, 8
Ex 5.3, 9
Ex 5.3, 10 (i)
Ex 5.3, 10 (ii) Important
Ex 5.3, 11 Important
Ex 5.3, 12
Ex 5.3, 13
Ex 5.3, 14 Important
Ex 5.3, 15 Deleted for CBSE Board 2022 Exams
Ex 5.3, 16 Important Deleted for CBSE Board 2022 Exams
Ex 5.3, 17 Deleted for CBSE Board 2022 Exams
Ex 5.3, 18 Important Deleted for CBSE Board 2022 Exams
Ex 5.3, 19 Important Deleted for CBSE Board 2022 Exams
Ex 5.3, 20 Important Deleted for CBSE Board 2022 Exams
Chapter 5 Class 10 Arithmetic Progressions (Term 2)
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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.
