An AP consists of 50 terms of which 3rd term is 12 and last term is

Ex 5.2, 8 - Chapter 5 Class 10 Arithmetic Progressions - Part 2
Ex 5.2, 8 - Chapter 5 Class 10 Arithmetic Progressions - Part 3

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Ex 5.2, 8 An A.P. consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term We know that an = a + (n – 1) d Given 3rd term is 12 a3 = a + (3 – 1) d 12 = a + 2d 12 – 2d = a a = 12 – 2d Given last term is 106 Last term = 50th term = a50 = 106 Now, a50 = a + (50 – 1) d 106 = a + 49 d 106 – 49d = a a = 106 – 49d From (1) and (2) 12 – 2d = 106 – 49d –2d + 49d = 106 – 12 47d = 94 d = 94/47 d = 2 Putting the value of d in (1) a = 12 – 2d a = 12 – 2 × 2 a = 12 – 4 a = 8 Now, We need to find 29th term i.e. a29 Now, an = a + (n – 1) d Putting values a29 = 8 + (29 – 1) × 2 = 8 + 28 × 2 = 8 + 56 = 64 Hence, 29th term = a29 = 64

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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