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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

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Ex 5.3, 3 In an AP (viii) Given an = 4, d = 2, Sn = βˆ’14, find n and a. Given an = 4, d = 2, Sn = –14 Since there are n terms, 𝑙 = an = 4 We use the formula Sn = 𝒏/𝟐 (𝒂+𝒍) Putting Sn = βˆ’14, 𝑙 = an = 4 –14 = 𝑛/2 (π‘Ž+4) –14 Γ— 2 =𝑛(π‘Ž+4) –28 = n (a + 4) (βˆ’28)/(π‘Ž + 4)=𝑛 n = (βˆ’πŸπŸ–)/(𝒂 + πŸ’) Also we know that an = a + (n – 1) d Putting an = 4 , d = 2 4 = a + (n – 1) Γ— 2 4 = a + 2n – 2 4 + 2 = a + 2n 6 = a + 2n Putting n = (βˆ’ πŸπŸ–)/(𝒂 + πŸ’) 6 = a + 2((βˆ’ 28)/(π‘Ž + 4)) 6 = a βˆ’ 56/(π‘Ž + 4) 6 = (π‘Ž(π‘Ž + 4) βˆ’ 56)/(π‘Ž + 4) 6 = (π‘Ž2 + 4π‘Ž βˆ’ 56)/(π‘Ž + 4) 6 (a + 4) = a2 + 4a – 56 6a + 24 = a2 4a – 56 0 = a2 + 4a – 56 – 6a – 24 0 = a2 – 2a – 80 a2 – 2a – 80 = 0 a2 – 10a + 8a– 80 = 0 a (a – 10) + 8 (a – 8) = 0 (a – 10) (a + 8) = 0 So, a = 10 or a = –8 Taking a = 10 n = (βˆ’ 28)/(π‘Ž + 4) n = (βˆ’ 28)/(10 + 4) n = (βˆ’ 28)/14 n = –2 Since n is number of terms, it cannot be negative So, n = –2 is not possible ∴ a = 10 is not possible Taking a = – 8 n = (βˆ’ 28)/(π‘Ž + 4) n = (βˆ’ 28)/(βˆ’ 8 + 4) n = (βˆ’ 28)/( βˆ’ 4) n = 7 So, n = 7

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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.