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

Ex 5.2, 11 You are here

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

Last updated at March 19, 2021 by Teachoo

Ex 5.2, 11 Which term of the A.P. 3, 15, 27, 39, … will be 132 more than its 54th term? Given AP 3, 15, 27, 39,… First we need to calculate 54th term Here, a = 3 d = 15 – 3 = 12 n = 54 Putting values in formula a54 = a + (n – 1) d = 3 + (54 – 1) × 12 = 3 + 53 × 12 = 3 + 636 = 639 So, 54th term = a54 = 639 We need to calculate term which is 132 more than 54th term i.e. should be 639 + 132 Therefore, an = 771 We need to find n Putting values in formula an = a + (n – 1) d 771 = 3 + (n – 1) × 12 771 – 3 = (n – 1) × 12 768 = (n – 1) × 12 768/12 = n – 1 64 = n – 1 n – 1 = 64 n = 64 + 1 n = 65 So, 65th term is 132 more than 54th term