Ex 5.3, 3 (v) - Chapter 5 Class 10 Arithmetic Progressions

Last updated at April 16, 2024 by

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

Ex 5.3, 3 - Chapter 5 Class 10 Arithmetic Progressions - Part 11
Ex 5.3, 3 - Chapter 5 Class 10 Arithmetic Progressions - Part 12

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Ex 5.3, 3 In an AP (v) Given d = 5, S9 = 75, find a and a9. Given d = 5 , S9 = 75 We use formula Sn = 𝒏/𝟐 (𝟐𝒂+(𝒏−𝟏)𝒅) Putting Sn = S9 = 75 , n = 9 , d = 5 75 = 9/2 (2𝑎+(9−1) × 5) 75 = 9/2 (2𝑎+8 × 5) 75 = 9/2(2𝑎+40) (75 × 2)/9 = 2a + 40 150/9 = 2a + 40 150 = 9 (2a + 40) 150 = 9 (2a) + 9(40) 150 = 18a + 360 150 – 360 = 18a –210 = 18a (−210)/18=𝑎 (−35)/3=𝑎 a = (−𝟑𝟓)/𝟑 Now, we need to find a9 We know that an = a + (n – 1) d Putting a = ( 35)/3, d = 5 & n = 9 a9 = (−35)/3 + (9 − 1) × 5 a9 = (−35)/3 + 8 × 5 a9 = (−35)/3 + 40 a9 = (−35 + 40 × 3)/3 a9 = (−35 + 120)/3 a9 = 𝟖𝟓/𝟑

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo