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 (Term 2)
  2. Serial order wise

Transcript

Ex 5.3, 4 How many terms of the AP. 9, 17, 25 … must be taken to give a sum of 636? Given AP is 9, 17, 25, ….. Here, a = 9 d = 17 – 9 = 8 & Sum = Sn = 636 We need to find n We know that Sum = 𝑛/2 (2a + (n – 1) d) Putting a = 9, d = 8, Sn = 636 636 = 𝑛/(2 ) (2 Γ— 9 + (n – 1) Γ— 8) 636 Γ— 2 = n (18 + 8n – 8) 1272 = n (10 + 8n) 1272 = 10n + 8n2 0 = 10n + 8n2 – 1272 10n + 8n2 – 1272 = 0 2(5n + 4n2 – 636)= 0 5n + 4n2 – 636 = 0 4n2 + 5n – 636 = 0 We solve by quadratic formula Comparing equation with ax2 + bx + c = 0 Here, a = 4, b = 5, c = – 636 We know that, D = b2 – 4ac D = (5)2 – 4 Γ— 4 Γ— (–636) D = 25 + 10176 D = 10201 Hence, roots to equation are given by n = (βˆ’ 𝒃 Β± βˆšπ‘«)/πŸπ’‚ Putting values n = (βˆ’ 5 Β± √𝟏𝟎𝟐𝟎𝟏)/(2 Γ— 4) n = (βˆ’ πŸ“ Β± 𝟏𝟎𝟏)/πŸ– Finding root of 10201 10201 = 101 Γ— 101 10201 = 1012 So, √("10201" ) = 101 Solving So, n = 12 & n = (βˆ’53)/4 Since n cannot be negative, ∴ n = 12 n = (βˆ’πŸ“ + 𝟏𝟎𝟏)/πŸ– n = 96/8 n = 12 n = (βˆ’πŸ“ βˆ’ 𝟏𝟎𝟏)/πŸ– n = (βˆ’106)/8 n = (βˆ’πŸ“πŸ‘)/π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.