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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 (2π‘Ž+(π‘›βˆ’1)𝑑) Putting a = 3, n = 8 , Sn = 192 192 = 8/2(2Γ—3+(8βˆ’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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