Ex 5.2, 3 (ii) - Chapter 5 Class 10 Arithmetic Progressions

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

Ex 5.2, 3 - Chapter 5 Class 10 Arithmetic Progressions - Part 4
Ex 5.2, 3 - Chapter 5 Class 10 Arithmetic Progressions - Part 5

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Transcript

Ex 5.2, 3 In the following APs find the missing term in the boxes (ii) ⎕, 13, ⎕, 3 Given a2 = 13 & a4 = 3 We know that an = a + (n – 1) d Given 2nd term is 13 a2 = a + (2 – 1) d 13 = a + d 13 – d = a a = 13 – d Given 4th term is 3 a4 = a + (4 – 1) d 3 = a + 3d 3 – 3d = a a = 3 – 3d From (1) & (2) 13 – d = 3 – 3d 13 – 3 = d – 3d 10 = –2d 10/(−2) = d – 5 = d d = – 5 Putting value of d in (1) a = 13 – d = 13 – ( – 5) = 13 + 5 = 18 Now, We need to find a3 a3 = a + (n – 1) d = 18 + (3 – 1) × – 5 = 18 + 2 × – 5 = 18 – 10 = 8 So, a3 = 8 Hence, , 13, , 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