Last updated at March 22, 2021 by Teachoo

Transcript

Ex 5.3, 8 Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively. We know that an = a + (n – 1) d From (1) & (2) 14 – d = 18 – 2d Given 2nd term is 14 a2 = a + (2 – 1) d 14 = a + d 14 – d = a a = 14 – d Given 3rd term is 18 a3 = a + (3 – 1) d 18 = a + 2d 18 – 2d = a a = 18 – 2d 2d – d = 18 – 14 d = 4 Putting value of d in (1) a = 14 – d a = 14 – 4 a = 10 Now, we need to find sum of first 51 terms We can use formula Sn = 𝒏/𝟐 [2a + (n – 1) d] Putting n = 51 , a = 10 & d = 4 S51 = 51/2 (2 × 10 + (51 − 1) × 4) S51 = 51/2 (20 + 50 × 4) S51 = 51/2 × (20 + 200) S51 = 51/2 × 220 S51 = 51 × 110 S51 = 5610 Hence, the sum of first 51 terms is 5610

Ex 5.3

Ex 5.3, 1

Ex 5.3, 2

Ex 5.3, 3

Ex 5.3, 4

Ex 5.3, 5

Ex 5.3, 6 Important

Ex 5.3, 7

Ex 5.3, 8 You are here

Ex 5.3, 9

Ex 5.3, 10 Important

Ex 5.3, 11 Important

Ex 5.3, 12

Ex 5.3, 13

Ex 5.3, 14

Ex 5.3, 15 Deleted for CBSE Board 2022 Exams

Ex 5.3, 16 Deleted for CBSE Board 2022 Exams

Ex 5.3, 17 Deleted for CBSE Board 2022 Exams

Ex 5.3, 18 Important Deleted for CBSE Board 2022 Exams

Ex 5.3, 19 Deleted for CBSE Board 2022 Exams

Ex 5.3, 20 Important Deleted for CBSE Board 2022 Exams

Chapter 5 Class 10 Arithmetic Progressions (Term 2)

Serial order wise

About the Author

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.