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

Ex 5.3, 3 - Chapter 5 Class 10 Arithmetic Progressions - Part 11
Ex 5.3, 3 - Chapter 5 Class 10 Arithmetic Progressions - Part 12

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

Transcript

Ex 5.3, 3 In an AP (v) Given d = 5, S9 = 75, find a and a9. Given d = 5 , S9 = 75 We use formula Sn = ๐’/๐Ÿ (๐Ÿ๐’‚+(๐’โˆ’๐Ÿ)๐’…) Putting Sn = S9 = 75 , n = 9 , d = 5 75 = 9/2 (2๐‘Ž+(9โˆ’1) ร— 5) 75 = 9/2 (2๐‘Ž+8 ร— 5) 75 = 9/2(2๐‘Ž+40) (75 ร— 2)/9 = 2a + 40 150/9 = 2a + 40 150 = 9 (2a + 40) 150 = 9 (2a) + 9(40) 150 = 18a + 360 150 โ€“ 360 = 18a โ€“210 = 18a (โˆ’210)/18=๐‘Ž (โˆ’35)/3=๐‘Ž a = (โˆ’๐Ÿ‘๐Ÿ“)/๐Ÿ‘ Now, we need to find a9 We know that an = a + (n โ€“ 1) d Putting a = ( 35)/3, d = 5 & n = 9 a9 = (โˆ’35)/3 + (9 โˆ’ 1) ร— 5 a9 = (โˆ’35)/3 + 8 ร— 5 a9 = (โˆ’35)/3 + 40 a9 = (โˆ’35 + 40 ร— 3)/3 a9 = (โˆ’35 + 120)/3 a9 = ๐Ÿ–๐Ÿ“/๐Ÿ‘

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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.