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Ex 5.1, 4 (xi) - (xv) - Which are APs? (xi) a, a2, a3, a4

Ex 5.1, 4 (xi) - (xv) - Chapter 5 Class 10 Arithmetic Progressions - Part 2

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Ex 5.1, 4 (xi) - (xv) - Chapter 5 Class 10 Arithmetic Progressions - Part 3

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Ex 5.1, 4 (xi) - (xv) - Chapter 5 Class 10 Arithmetic Progressions - Part 7

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Ex 5.1, 4 (xi) - (xv) - Chapter 5 Class 10 Arithmetic Progressions - Part 9

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Ex 5.1, 4 (xi) - (xv) - Chapter 5 Class 10 Arithmetic Progressions - Part 11

Ex 5.1, 4 (xi) - (xv) - Chapter 5 Class 10 Arithmetic Progressions - Part 12
Ex 5.1, 4 (xi) - (xv) - Chapter 5 Class 10 Arithmetic Progressions - Part 13

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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. (xi) a, a2, a3, a4 … a, a2 , a3 , a4 … Difference between second and first term = a2 – a = a (a – 1) Difference between third and second term = a3 – a2 = a2 (a – 1) Since, difference is not same. a (a – 1) ≠ a2 (a – 1) 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. (xii) √2, √8, √18, √32……… √2 , √8 , √18 , √32 Difference between second and first term = √8 – √2 = √(2 × 2 × 2) – √2 = √(2 × 22) – √2 = 2√2 – √2 = √𝟐 Difference between third and second term = √18 – √8 = √(3 × 3 × 2) – √(2 × 2 × 2) = √(32 × 2) – √(22 × 2) = 3 √2 – 2 √2 = √𝟐 Difference between fourth and third term = √32 – √18 = √(4 × 4 × 2) – √(2 × 3 × 3) = √(42 × 2) – √(32 × 2) = 4 √2 – 3 √2 = √𝟐 Since difference is same, it is an AP Common difference = d = √𝟐 We have to find next 3 terms, We are given 4 terms So, we need to find 5th, 6th and 7th term Fifth term = fourth term + common difference = √32 + √2 = √(4 × 4 × 2) + √2 = √(42 × 2) + √2 = 4 √2 + √2 = 5 √𝟐 Sixth term = Fifth term + Common difference = 5√2 + √2 = 6√𝟐 Seventh term = Sixth term + common difference = 6 √2 + √2 = 7√𝟐 Hence, 5th, 6th and 7th terms are 5√𝟐 , 6√𝟐, 7√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. (xiii) √3, √6, √9, √12……… √3 , √6 , √9 , √12 Difference between second and first term = √6 – √3 = √(3 × 2) – √3 = √3 × √2 – √3 = √3 × √2 – √3 × 1 = √𝟑 (√𝟐 – 1) Difference between third and second term = √9 – √6 = √(3 × 3) – √(3 × 2) = √3 × √3 – √3 × √2 = √𝟑 (√𝟑 – √𝟐) The difference is not same √3 (√2 – 1) ≠ √3 (√3 – √2) So, it is not A.P. 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 (xiv) 12, 32, 52, 72 … 12, 32, 52, 72 Difference between second and first term = 32 – 12 = 9 – 1 = 8 Difference between third and second term = 52 – 32 = 25 – 9 = 16 Difference is not same 8 ≠ 16 So, it is not A.P 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 (xv) 12, 52, 72, 73 … 12, 52, 72, 73 Difference between second and first term = 52 – 12 = 25 – 1 = 24 Difference between third and second term = 72 – 52 = 49 – 25 = 24 Difference between fourth and third term = 73 – 49 = 24 Since difference is same, it is an AP. Common difference = d = 24 We have to find next 3 terms, We are given 4 terms So, we need to find 5th, 6th and 7th term Fifth term = fourth term + common difference = 73 + 24 = 97 Sixth term = Fifth term + Common Difference = 97 + 24 = 121 Seventh term = Sixth term + Common Difference = 121 + 24 = 145 So, next 3 terms of AP are 97, 121 and 145

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.