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Last updated at May 29, 2018 by Teachoo

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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. (i) 2, 4, 8, 16 2, 4, 8, 16 .. Difference of second and first term = 4 2 = 2 Difference of third and second term = 8 4 = 4 Since 2 4 Difference is not same 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. (ii) 2, 5/2, 3, 7/2 .. 2, 5/2 , 3 , 7/2 , .. Difference between second and first term = 5/2 2 = (5 2 2)/2 = (5 4)/2 = 1/2 Difference between third and second term = 3 5/2 = (3 2 5)/2 = (3 2 5)/2 = (6 5)/2 = 1/2 Difference between fourth and third term = 7/2 3 = (7 3 2)/2 = (7 6)/2 = 1/2 Since difference is same, it is an AP Common difference = d = 1/2 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 = 7/2 + 1/2 = (7 + 1)/2 = 8/2 = 4 6th term = 5th term + common difference = 4 + 1/2 = (4 2 + 1)/2 = 9/2 7th term = 6th term + common difference = 9/2 + 1/2 = (9 + 1)/2 = 10/2 = 5 Hence 5th , 6th and 7th terms are 4, 9/2 , 5 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. (iii) 1.2, 3.2, 5.2, 7.2 1.2, 3.2, 5.2, 7.2 Difference between second and first term = 3.2 ( 1.2) = 3.2 + 1.2 = 2.0 = 2 Difference between third and second term = 5.2 ( 3.2) = 5.2 + 3.2 = 2.0 = 2 Difference between fourth and third term = 7.2 5.2 = 7.2 + 5.2 = 2.0 = 2 Since difference is same, it is an AP Common difference = d = 2 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 = 7.2 + 2 = 7.2 2 = 9.2 6th term = fifth term + common difference = 9.2 + 2 = 9.2 2 = 11.2 7th term = sixth term + common difference = 11.2 + 2 = 11.2 2 = 13.2 Hence 5th , 6th and 7th terms are 9.2, 11.2, 13.2 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. (iv) 10, 6, 2, 2 10, 6, 2, 2, . Difference between 2nd and 1st term = 6 ( 10) = 6 + 10 = 4 Difference between third and second term = 2 ( 6) = 2 + 6 = 4 Difference between fourth and third term = 2 ( 2) = 2 + 2 = 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 = fourth term + common difference = 2 + 4 = 6 6th term = fifth term + common difference = 6 + 4 = 10 7th term = sixth term + common difference = 10 + 4 = 14 Hence 5th , 6th and 7th terms are 6, 10 , 14 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. (v) 3,3 + 2, 3 + 2 2,3 + 3 2 .. 3, 3 + 2 , 3 + 2 2 , 3 + 3 2 , . Difference between second and first term = 3 + 2 3 = 2 Difference between third and second term = (3 + 2 2 ) (3 + 2 ) = 3 + 2 2 3 2 = 3 3 + 2 2 2 = 0 + 2 (2 1) = 2 (1) = 2 Difference between fourth and third term = (3 + 3 2 ) (3 + 2 2) = 3 + 3 2 3 2 2 = 3 3 + 3 2 2 2 = 0 + 3 2 2 2 = 2 (3 2 ) = 2 1 = 2 Since difference is same, it is an AP Common difference = d = 2 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 = (3 + 3 2) + 2 = 3 + 2 [ 3 + 1] = 3 + 2 (4) = 3 + 4 2 6th term = 5th term + common difference = (3 + 4 2) + 2 = 3 + 2 (4 + 1) = 3 + 2 5 = 3 + 5 2 Similarly ,7th term = 6th term + common difference = (3 + 5 2) + 2 = 3 + 2 (5 + 1) = 3 + 2 6 = 3 + 6 2

Checking if AP or not and finding a, d

Chapter 5 Class 10 Arithmetic Progressions

Concept wise

- Checking if AP or not and finding a, d
- Finding nth term
- Finding n
- Finding AP
- Finding nth term from end
- Divisible/ multiples
- Statement questions - nth term
- Finding sum of n terms
- Finding number of terms given s
- Given nth term finding s
- Statement questions - Sum of n terms
- Determining AP and finding sum

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.