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

  1. Chapter 5 Class 10 Arithmetic Progressions (Term 2)
  2. Serial order wise

Transcript

Ex 5.3, 3 (iv) Given a3 = 15, S10 = 125, find d and a10. For a3 = 15 We know that an = a + (n โ€“ 1) d Putting an = a3 = 15 & n = 3 15 = a + 2d 15 โ€“ 2d = a a = 15 โ€“ 2d For S10 = 125 We know that Sn = ๐‘›/2(๐‘Ž+(๐‘›โˆ’1)๐‘‘) Putting Sn = S10 = 125, n = 10 S10 = 10/2 (2๐‘Ž+(10โˆ’1)๐‘‘) 125 = 5 (2a + 9d) 125/5=2๐‘Ž+9๐‘‘ 25 = 2a + 9d 25 โ€“ 9d = 2a a = (๐Ÿ๐Ÿ“ โˆ’ ๐Ÿ—๐’…)/๐Ÿ From (1) and (2) 15 โ€“ 2d = (๐Ÿ๐Ÿ‘โˆ’๐Ÿ—๐’…)/๐Ÿ 30 โ€“ 4d = 25 โ€“ 9d 5 = โ€“ 5d 5/(โˆ’5)=๐‘‘ d = โ€“1 Putting value of d in (1) a = 15 โ€“ 2d a = 15 โ€“ 2 ร—(โˆ’1) a = 15 + 2 a = 17 Now we need to find a10 an = a + (n โ€“ 1) d Putting n = 10 , a = 17 & d = โ€“1 a10 = 17 + (10 โ€“ 1) ร— (โˆ’1) a10 = 17 + 9 ร— (โˆ’1) a10 = 17 โ€“ 9 a10 = 8

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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.