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Ex 5.2, 3 - Chapter 5 Class 10 Arithmetic Progressions - Part 6

Ex 5.2, 3 - Chapter 5 Class 10 Arithmetic Progressions - Part 7
Ex 5.2, 3 - Chapter 5 Class 10 Arithmetic Progressions - Part 8 Ex 5.2, 3 - Chapter 5 Class 10 Arithmetic Progressions - Part 9

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Ex 5.2, 3 In the following APs find the missing term in the boxes (iii) 5, ⎕, ⎕, 9 1/2 Given a = 5 , a4 = 91/2 = (2(9) + 1)/2 = 19/2 We know that an = a + (n – 1) d Putting values a4 = a + (4 – 1) d 19/2 = 5 + 3d 19/2 – 5 = 3d (19 − 2(5))/2 = 3d 9/2 = 3d 9/2 × 1/3 = d 3/2 = d d = 𝟑/𝟐 Hence, a = 5 and d = 3/2 Now, We need to find a2 & a3 Hence, a2 = 13/2 = 61/2 a3 = 8 a2 an = a + (n – 1) d a2 = a + (2 – 1)d = a + d = 5 + 3/2 = (5(2) + 3)/2 = 𝟏𝟑/𝟐 a3 an = a + (n – 1) d a3 = a + (3 – 1)d = a + 2d = 5 + 2(3/2) = 5 + 3 = 8 So, 5, , , 91/2

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.