Ex 5.2, 3 - In the following APs, find missing term in boxes

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

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

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

Ex 5.2, 3 - Chapter 5 Class 10 Arithmetic Progressions - Part 10 Ex 5.2, 3 - Chapter 5 Class 10 Arithmetic Progressions - Part 11 Ex 5.2, 3 - Chapter 5 Class 10 Arithmetic Progressions - Part 12

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

  1. Chapter 5 Class 10 Arithmetic Progressions
  2. Serial order wise

Transcript

Ex 5.2, 3 In the following APs find the missing term in the boxes (i) 2, ⎕ ,26 Let 2nd term = x So, AP is 2, x, 26 Since it is an AP, Common difference is same ∴ Second term – First term = Third term – Second term x – 2 = 26 – x x + x = 2 + 26 2x = 28 x = 28/2 x = 14 So, 2, ,26 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 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 Ex 5.2, 3 In the following APs find the missing term in the boxes (iv) – 4, ⎕, ⎕, ⎕, ⎕, 6 Given, a = –4, a6 = 6 We know that an = a + (n – 1) d Putting values a6 = – 4 + (6 – 1) × d 6 = – 4 + 5 d 6 + 4 = 5d 10 = 5d 10/5 = d 2 = d d = 2 Now , a = –4, d = 2 We need to find a2, a3, a4, a5 a2 a2 = a + (2 – 1) d = a + d = –4 + 2 = –2 a3 a3 = a + (3 – 1) d = a + 2d = –4 + 2(2) = 0 a4 a4 = a + (4 – 1) d = a + 3d = –4 + 3(2) = 2 a5 a3 = a + (5 – 1)d = a + 4d = –4 + 4(2) = 4 Hence, – 4, , , , , 6 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
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.