Ex 5.3, 9 - If sum of first 7 terms of an AP is 49, and 17 term is 289

Ex 5.3, 9 - Chapter 5 Class 10 Arithmetic Progressions - Part 2
Ex 5.3, 9 - Chapter 5 Class 10 Arithmetic Progressions - Part 3

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

Transcript

Ex 5.3, 9 If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms. We know that Sn = ๐‘›/2 (2๐‘Ž+(๐‘›โˆ’1)๐‘‘) Sum of first 7 terms = 49 S7 = 7/2 (2๐‘Ž+(๐‘›โˆ’1)๐‘‘) 49 = 7/2 (2๐‘Ž+(7โˆ’1)๐‘‘) 49 = 7/2 (2๐‘Ž+6๐‘‘) (49 ร— 2)/7 "= 2a + 6d" "14 = 2a + 6d" (14 โˆ’ 6๐‘‘)/2=๐‘Ž a = 7 โ€“ 3d Sum of first 17 terms = 289 S17 = 17/2 (2๐‘Ž+(17โˆ’1)๐‘‘) 289 = 17/2 (2a + (17 โ€“ 1) d) 289 = 17/2 (2a + 16d) (289 ร— 2)/17 = 2a + 16d 34 = 2a + 16 d (34 โˆ’ 16๐‘‘)/2 = a a = 17 โ€“ 8d From (1) and (2) 7 โ€“ 3d = 17 โ€“ 8d 8d โ€“ 3d = 17 โ€“ 7 5d = 10 d = 10/5 d = 2 Putting value of d in (1) a = 7 โ€“ 3d a = 7 โ€“ 3 ร—2 a = 7 โ€“ 6 a = 1 Hence, a = 1 & d = 2 We need to find sum of first n terms We can use formula Sn = ๐’/๐Ÿ (2a + (n โ€“ 1) d) Putting a = 1 & d = 2 = ๐‘›/2 (2 ร— 1+(๐‘›โˆ’1)2) = ๐‘›/2(2+2๐‘›โˆ’2) = ๐‘›/2 (0 + 2n) = ๐‘›/2 ร— 2n = n2

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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.