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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

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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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.