Ex 5.3, 3 (iv) - Chapter 5 Class 10 Arithmetic Progressions
Last updated at Dec. 16, 2024 by Teachoo
Last updated at Dec. 16, 2024 by Teachoo
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(𝑎+(𝑛−1)𝑑) Putting Sn = S10 = 125, n = 10 S10 = 10/2 (2𝑎+(10−1)𝑑) 125 = 5 (2a + 9d) 125/5=2𝑎+9𝑑 25 = 2a + 9d 25 – 9d = 2a a = (𝟐𝟓 − 𝟗𝒅)/𝟐 From (1) and (2) 15 – 2d = (𝟐𝟑−𝟗𝒅)/𝟐 30 – 4d = 25 – 9d 5 = – 5d 5/(−5)=𝑑 d = –1 Putting value of d in (1) a = 15 – 2d a = 15 – 2 ×(−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) × (−1) a10 = 17 + 9 × (−1) a10 = 17 – 9 a10 = 8
Ex 5.3
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 You are here
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)
Ex 5.3, 3 (x)
Ex 5.3, 4
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Ex 5.3, 10 (i)
Ex 5.3, 10 (ii) Important
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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