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Ex 5.3, 3 (ix) - Chapter 5 Class 10 Arithmetic Progressions
Last updated at May 29, 2023 by Teachoo
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
Ex 5.3, 16 Important
Ex 5.3, 17
Ex 5.3, 18 Important
Ex 5.3, 19 Important
Ex 5.3, 20 Important
Chapter 5 Class 10 Arithmetic Progressions
Serial order wise
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 = 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
