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Ex 5.3, 3 - Chapter 5 Class 10 Arithmetic Progressions - Part 7

Ex 5.3, 3 - Chapter 5 Class 10 Arithmetic Progressions - Part 8
Ex 5.3, 3 - Chapter 5 Class 10 Arithmetic Progressions - Part 9

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Ex 5.3, 3 (iv) Given a3 = 15, S10 = 125, find d and a10. For a3 = 15 We know that an = a + (n – 1) d Putting an = a3 = 15 & n = 3 15 = a + 2d 15 – 2d = a a = 15 – 2d For S10 = 125 We know that Sn = 𝑛/2(π‘Ž+(π‘›βˆ’1)𝑑) Putting Sn = S10 = 125, n = 10 S10 = 10/2 (2π‘Ž+(10βˆ’1)𝑑) 125 = 5 (2a + 9d) 125/5=2π‘Ž+9𝑑 25 = 2a + 9d 25 – 9d = 2a a = (πŸπŸ“ βˆ’ πŸ—π’…)/𝟐 From (1) and (2) 15 – 2d = (πŸπŸ‘βˆ’πŸ—π’…)/𝟐 30 – 4d = 25 – 9d 5 = – 5d 5/(βˆ’5)=𝑑 d = –1 Putting value of d in (1) a = 15 – 2d a = 15 – 2 Γ—(βˆ’1) a = 15 + 2 a = 17 Now we need to find a10 an = a + (n – 1) d Putting n = 10 , a = 17 & d = –1 a10 = 17 + (10 – 1) Γ— (βˆ’1) a10 = 17 + 9 Γ— (βˆ’1) a10 = 17 – 9 a10 = 8

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