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) You are here

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

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, 3 In an AP (iii) Given a12 = 37, d = 3, find a and S12. Given a12 = 37, d = 3 Finding a We know that an = a + (n – 1) d Putting d = 3, n = 12 and a12 = 37 37 = a + (12 – 1) × 3 37 = a + 11 × 3 37 = a + 33 37 − 33 = a 4 = a a = 4 Now, we can find (S12) by using formula Sn = 𝒏/𝟐 (𝒂+𝒍) Putting n = 12 , a = 4, 𝑙 = a12 = 37 S12 = 12/2 (4+37) = 12/2×41 = 6 ×41 = 246