Ex 5.2, 3 - Chapter 5 Class 10 Arithmetic Progressions - Part 13

Ex 5.2, 3 - Chapter 5 Class 10 Arithmetic Progressions - Part 14
Ex 5.2, 3 - Chapter 5 Class 10 Arithmetic Progressions - Part 15
Ex 5.2, 3 - Chapter 5 Class 10 Arithmetic Progressions - Part 16

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Transcript

Ex 5.2, 3 In the following APs find the missing term in the boxes (v) ⎕, 38, ⎕ , ⎕, ⎕, − 22 Given a2 = 38 & a6 = – 22 We know that an = a + (n – 1) d Given 2nd term is 38 a2 = a + (2 – 1) d 38 = a + d 38 – d = a a = 38 – d Given 6th term is –22 a4 = a + (6 – 1) d –22 = a + 5d –22 – 5d = a a = –22 – 5d From (1) & (2) 38 – d = –22 – 5d 38 + 22 = d – 5d 60 = –4d 60/(−4) = d –15 = d d = –15 Putting value of d in (1) a = 38 – d = 38 – ( – 15) = 38 + 15 = 53 Hence, a = 53, d = –15 We need to find a3, a4, a5 We know that an = a + (n – 1)d a3 a3 = a + (3 – 1) d = a + 2d = 53 + 2(−15) = 53 − 30 = 23 a4 a4 = a + (4 – 1) d = a + 3d = 53 + 3(−15) = 53 − 45 = 8 a5 a3 = a + (5 – 1)d = a + 4d = 53 + 4(−15) = 53 − 60 = −7 Hence, ,38, , , , - 22

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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