Ex 5.3, 3 - Chapter 5 Class 10 Arithmetic Progressions - Part 19

Ex 5.3, 3 - Chapter 5 Class 10 Arithmetic Progressions - Part 20
Ex 5.3, 3 - Chapter 5 Class 10 Arithmetic Progressions - Part 21
Ex 5.3, 3 - Chapter 5 Class 10 Arithmetic Progressions - Part 22

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

Transcript

Ex 5.3, 3 In an AP (viii) Given an = 4, d = 2, Sn = โˆ’14, find n and a. Given an = 4, d = 2, Sn = โ€“14 Since there are n terms, ๐‘™ = an = 4 We use the formula Sn = ๐’/๐Ÿ (๐’‚+๐’) Putting Sn = โˆ’14, ๐‘™ = an = 4 โ€“14 = ๐‘›/2 (๐‘Ž+4) โ€“14 ร— 2 =๐‘›(๐‘Ž+4) โ€“28 = n (a + 4) (โˆ’28)/(๐‘Ž + 4)=๐‘› n = (โˆ’๐Ÿ๐Ÿ–)/(๐’‚ + ๐Ÿ’) Also we know that an = a + (n โ€“ 1) d Putting an = 4 , d = 2 4 = a + (n โ€“ 1) ร— 2 4 = a + 2n โ€“ 2 4 + 2 = a + 2n 6 = a + 2n Putting n = (โˆ’ ๐Ÿ๐Ÿ–)/(๐’‚ + ๐Ÿ’) 6 = a + 2((โˆ’ 28)/(๐‘Ž + 4)) 6 = a โˆ’ 56/(๐‘Ž + 4) 6 = (๐‘Ž(๐‘Ž + 4) โˆ’ 56)/(๐‘Ž + 4) 6 = (๐‘Ž2 + 4๐‘Ž โˆ’ 56)/(๐‘Ž + 4) 6 (a + 4) = a2 + 4a โ€“ 56 6a + 24 = a2 4a โ€“ 56 0 = a2 + 4a โ€“ 56 โ€“ 6a โ€“ 24 0 = a2 โ€“ 2a โ€“ 80 a2 โ€“ 2a โ€“ 80 = 0 a2 โ€“ 10a + 8aโ€“ 80 = 0 a (a โ€“ 10) + 8 (a โ€“ 8) = 0 (a โ€“ 10) (a + 8) = 0 So, a = 10 or a = โ€“8 Taking a = 10 n = (โˆ’ 28)/(๐‘Ž + 4) n = (โˆ’ 28)/(10 + 4) n = (โˆ’ 28)/14 n = โ€“2 Since n is number of terms, it cannot be negative So, n = โ€“2 is not possible โˆด a = 10 is not possible Taking a = โ€“ 8 n = (โˆ’ 28)/(๐‘Ž + 4) n = (โˆ’ 28)/(โˆ’ 8 + 4) n = (โˆ’ 28)/( โˆ’ 4) n = 7 So, n = 7

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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.