Ex 5.3, 4 - How many terms of AP 9, 17, 25 … must be - Ex 5.3

Ex 5.3, 4 - Chapter 5 Class 10 Arithmetic Progressions - Part 2
Ex 5.3, 4 - Chapter 5 Class 10 Arithmetic Progressions - Part 3 Ex 5.3, 4 - Chapter 5 Class 10 Arithmetic Progressions - Part 4

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

Transcript

Ex 5.3, 4 How many terms of the AP. 9, 17, 25 โ€ฆ must be taken to give a sum of 636? Given AP is 9, 17, 25, โ€ฆ.. Here, a = 9 d = 17 โ€“ 9 = 8 & Sum = Sn = 636 We need to find n We know that Sum = ๐‘›/2 (2a + (n โ€“ 1) d) Putting a = 9, d = 8, Sn = 636 636 = ๐‘›/(2 ) (2 ร— 9 + (n โ€“ 1) ร— 8) 636 ร— 2 = n (18 + 8n โ€“ 8) 1272 = n (10 + 8n) 1272 = 10n + 8n2 0 = 10n + 8n2 โ€“ 1272 10n + 8n2 โ€“ 1272 = 0 2(5n + 4n2 โ€“ 636)= 0 5n + 4n2 โ€“ 636 = 0 4n2 + 5n โ€“ 636 = 0 We solve by quadratic formula Comparing equation with ax2 + bx + c = 0 Here, a = 4, b = 5, c = โ€“ 636 We know that, D = b2 โ€“ 4ac D = (5)2 โ€“ 4 ร— 4 ร— (โ€“636) D = 25 + 10176 D = 10201 Hence, roots to equation are given by n = (โˆ’ ๐’ƒ ยฑ โˆš๐‘ซ)/๐Ÿ๐’‚ Putting values n = (โˆ’ 5 ยฑ โˆš๐Ÿ๐ŸŽ๐Ÿ๐ŸŽ๐Ÿ)/(2 ร— 4) n = (โˆ’ ๐Ÿ“ ยฑ ๐Ÿ๐ŸŽ๐Ÿ)/๐Ÿ– Finding root of 10201 10201 = 101 ร— 101 10201 = 1012 So, โˆš("10201" ) = 101 Solving So, n = 12 & n = (โˆ’53)/4 Since n cannot be negative, โˆด n = 12 n = (โˆ’๐Ÿ“ + ๐Ÿ๐ŸŽ๐Ÿ)/๐Ÿ– n = 96/8 n = 12 n = (โˆ’๐Ÿ“ โˆ’ ๐Ÿ๐ŸŽ๐Ÿ)/๐Ÿ– n = (โˆ’106)/8 n = (โˆ’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.