Last updated at March 19, 2021 by Teachoo

Transcript

Ex 5.2, 10 If 17th term of an A.P. exceeds its 10th term by 7. Find the common difference. We know that an = a + (n – 1) d So, a17 = a + (17 – 1) d a17 = a + 16d Also, a10 = a + (10 – 1)d a10 = a + 9d Given that 17th term of an A.P. exceeds its 10th term by 7 a17 – a10 = 7 (a + 16d) – (a + 9d) = 7 a + 16d – a – 9d = 7 a – a + 16d – 9d = 7 0 + 7d = 7 7d = 7 7d = 7 d = 7/7 d = 1 Therefore, common difference of the given AP is 1 (From (1) & (2))

Ex 5.2

Ex 5.2, 1

Ex 5.2, 2 (i) (MCQ)

Ex 5.2, 2 (ii) (MCQ) Important

Ex 5.2, 3 (i)

Ex 5.2, 3 (ii)

Ex 5.2, 3 (iii) Important

Ex 5.2, 3 (iv)

Ex 5.2, 3 (v) Important

Ex 5.2, 4

Ex 5.2, 5 (i)

Ex 5.2, 5 (ii) Important

Ex 5.2, 6 Important

Ex 5.2, 7

Ex 5.2, 8

Ex 5.2, 9

Ex 5.2, 10 You are here

Ex 5.2, 11

Ex 5.2, 12 Important

Ex 5.2, 13 Important

Ex 5.2, 14 Important

Ex 5.2, 15

Ex 5.2, 16

Ex 5.2, 17 Important

Ex 5.2, 18

Ex 5.2, 19

Ex 5.2, 20 Important

Chapter 5 Class 10 Arithmetic Progressions (Term 2)

Serial order wise

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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.