Ex 5.3, 3 - Chapter 5 Class 10 Arithmetic Progressions - Part 13

Ex 5.3, 3 - Chapter 5 Class 10 Arithmetic Progressions - Part 14
Ex 5.3, 3 - Chapter 5 Class 10 Arithmetic Progressions - Part 15
Ex 5.3, 3 - Chapter 5 Class 10 Arithmetic Progressions - Part 16

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

Transcript

Ex 5.3, 3 In an AP (vi) Given a = 2, d = 8, Sn = 90, find n and an. Given a = 2, d = 8, Sn = 90 We can use formula Sn = ๐‘›/2 (2๐‘Ž+(๐‘›โˆ’1)๐‘‘) Putting a = 2, d = 8, Sn = 90 90 = ๐‘›/2 (2 ร— 2+(๐‘›โˆ’1) ร— 8) 90 = ๐‘›/2 (4+8๐‘›โˆ’8) 90 ร— 2 =๐‘›(4+8๐‘›โˆ’8) 180 = n (8n โ€“ 4) 180 = 8n2 โ€“ 4n โ€“180 + 8n2 โ€“ 4n = 0 8n2 โ€“ 4n โ€“ 180 = 0 8n2 โ€“ 4n โ€“ 180 = 0 4 (2n2 โ€“ n โ€“ 45) = 0 2n2 โ€“ n โ€“ 45 = 0 2n2 โ€“ 10n + 9n โ€“ 45 = 0 2n(n โ€“ 5) + 9(n โ€“ 5) = 0 (2n + 9) (n โˆ’ 5) = 0 2n + 9 = 0 2n = โ€“9 n = (โˆ’๐Ÿ—)/๐Ÿ n โ€“ 5 = 0 n = 5 Therefore, n = 5 & n = (โˆ’9)/2 But n cannot be in fraction, So, n = 5 We need to find an, i.e. a5 We know that an = a + (n โ€“ 1)d Putting a = 2, n = 5, d = 8 a5 = 2 + (5 โ€“ 1) 8 a5 = 2 + (4)8 a5 = 2 + 32 a5 = 34 Therefore, n = 5 & a5 = 34

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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.