Ex 5.3

Ex 5.3, 1 (i)

Ex 5.3, 1 (ii)

Ex 5.3, 1 (iii) Important

Ex 5.3, 1 (iv)

Ex 5.3, 2 (i)

Ex 5.3, 2 (ii)

Ex 5.3, 2 (iii) Important

Ex 5.3, 3 (i)

Ex 5.3, 3 (ii)

Ex 5.3, 3 (iii)

Ex 5.3, 3 (iv) Important

Ex 5.3, 3 (v)

Ex 5.3, 3 (vi) Important

Ex 5.3, 3 (vii)

Ex 5.3, 3 (viii) Important

Ex 5.3, 3 (ix)

Ex 5.3, 3 (x)

Ex 5.3, 4

Ex 5.3, 5

Ex 5.3, 6 Important

Ex 5.3, 7

Ex 5.3, 8

Ex 5.3, 9

Ex 5.3, 10 (i)

Ex 5.3, 10 (ii) Important You are here

Ex 5.3, 11 Important

Ex 5.3, 12

Ex 5.3, 13

Ex 5.3, 14 Important

Ex 5.3, 15 Deleted for CBSE Board 2022 Exams

Ex 5.3, 16 Important 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 Important 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

Last updated at Aug. 3, 2021 by Teachoo

Ex 5.3, 10 Show that a1, a2 … , an , … form an AP where an is defined as below (ii) an = 9 − 5n . Also find the sum of first 15 terms in each case Since an = 9 – 5n Hence, the series is 4, –1, –6, …… Since difference is same, it is an AP Common difference = d = –1 – 4 = –5 Taking n = 1 a1 = 9 – 5 × 1 a = 9 – 5 a = 4 Taking n = 2 a2 = 9 – 5 × 2 a2 = 9 – 10 a2 = –1 Taking n = 3, a3 = 9 – 5 × 3 a3 = 9 – 15 a3 = –6 Now we have to find sum of first 15 terms i.e. S15 We know that Sn = 𝒏/𝟐 (𝟐𝒂+(𝒏−𝟏)𝒅) Putting n = 15, a = 4, d = −5 S15 = 15/2 (2 × 4+(15−1)×−5) S15 = 15/2 (8+14×−5) S15 = 15/2 (8−70) S15 = 15/2× − 62 S15 = –465