Ex 5.1, 4 (vi) - (x) - Which are APs? 0.2, 0.22, 0.222 - Ex 5.1

Ex 5.1, 4 (vi) - (x) - Chapter 5 Class 10 Arithmetic Progressions - Part 2

Ex 5.1, 4 (vi) - (x) - Chapter 5 Class 10 Arithmetic Progressions - Part 3

Ex 5.1, 4 (vi) - (x) - Chapter 5 Class 10 Arithmetic Progressions - Part 4
Ex 5.1, 4 (vi) - (x) - Chapter 5 Class 10 Arithmetic Progressions - Part 5

Ex 5.1, 4 (vi) - (x) - Chapter 5 Class 10 Arithmetic Progressions - Part 6 Ex 5.1, 4 (vi) - (x) - Chapter 5 Class 10 Arithmetic Progressions - Part 7 Ex 5.1, 4 (vi) - (x) - Chapter 5 Class 10 Arithmetic Progressions - Part 8 Ex 5.1, 4 (vi) - (x) - Chapter 5 Class 10 Arithmetic Progressions - Part 9

Ex 5.1, 4 (vi) - (x) - Chapter 5 Class 10 Arithmetic Progressions - Part 10

Ex 5.1, 4 (vi) - (x) - Chapter 5 Class 10 Arithmetic Progressions - Part 11 Ex 5.1, 4 (vi) - (x) - Chapter 5 Class 10 Arithmetic Progressions - Part 12 Ex 5.1, 4 (vi) - (x) - Chapter 5 Class 10 Arithmetic Progressions - Part 13

Go Ad-free

Transcript

Ex 5.1, 4 Which of the following are APs? If they form an A.P. find the common difference d and write three more terms. (vi) 0.2, 0.22, 0.222, 0.2222 …. 0.2, 0.22, 0.222, 0.2222 Difference between second and first term = 0.22 – 0.2 = 0.22 – 0.20 = 0.02 Difference between third and second term = 0.222 – 0.22 = 0.222 – 0.220 = 0.002 Since difference is not same 0.02 ≠ 0.002 Hence, it is not an AP Ex 5.1, 4 Which of the following are APs? If they form an A.P. find the common difference d and write three more terms. (vii) 0, − 4, − 8, − 12 … 0, –4, –8, –12 Difference between second and first term = ( – 4) – 0 = –4 Difference between third and second term = –8 – (– 4) = –8 + 4 = –4 Difference between fourth and third term = –12 – (–8) = –12 + 8 = –4 Since difference is same, it is an AP Common difference = d = –4 We have to find next three terms We are given 4 terms . So, we have to find 5th, 6th & 7th terms. 5th term = 4th term + common difference = – 12 + (–4) = – 12 – 4 = –16 6th term = 5th term + common difference = – 16 + (–4) = –20 7th term = 6th term + common difference = –20 + (–4) = –24 Hence, 5th 6th and 7th term are –16, –20, –24 Ex 5.1, 4 Which of the following are APs? If they form an A.P. find the common difference d and write three more terms. (viii) (−1)/2 , (−1)/2 , (−1)/2 , (−1)/2 … (−1)/2 , (−1)/2 , (−1)/2 , (−1)/2 … All terms are same Difference between second and first term = ((− 1)/2) −" " ((− 1)/2) = (−1)/2 + 1/2 = 0 Similarly difference between third and second term = ((− 1)/2) −" " ((− 1)/2) = (−1)/2 + 1/2 = 0 Difference between fourth and third term = ((− 1)/2) −" " ((− 1)/2) = (−1)/2 + 1/2 = 0 Since difference is same, it is an AP Common difference = d = 0 We have to find next three terms We are given 4 terms . So, we have to find 5th, 6th & 7th terms 5th term = fourth term + common difference = (−1)/2 + 0 = (−𝟏)/𝟐 6th term = 5th Term + Common difference = (−1)/2 + 0 = (−𝟏)/𝟐 7th term = 6th Term + Common difference = (−1)/2 + 0 = (−𝟏)/𝟐 Therefore, 5th , 6th and 7th terms are (−𝟏)/𝟐 , (−𝟏)/𝟐 , (−𝟏)/𝟐 Ex 5.1, 4 Which of the following are APs? If they form an A.P. find the common difference d and write three more terms. (ix) 1, 3, 9, 27 … 1, 3, 9, 27 Difference between second and first term = 3 – 1 = 2 Difference between third and second term = 9 – 3 = 6 Since, difference is not same 2 ≠ 6 Hence it is not an AP Ex 5.1, 4 Which of the following are APs? If they form an A.P. find the common difference d and write three more terms. (x) a, 2a, 3a, 4a … a, 2a, 3a, 4a Difference between second and first term = 2a – a = a Difference between third and second term = 3a – 2a = a Difference between fourth and third term = 4a – 3a = a Since difference is same, it is an AP Common difference = d = a We have to find next three terms We are given 4 terms . So, we have to find 5th , 6th & 7th terms 5th term = fourth term + common difference = 4a + a = 5a 6th term = fifth term + common difference = 5a + a = 6a Seventh term = sixth term + common difference = 6a + a = 7a Hence, 5th, 6th and 7th terms are 5a , 6a, 7a

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

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, Social Science, Physics, Chemistry, Computer Science at Teachoo.