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) You are here

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

Ex 5.3, 16 Important

Ex 5.3, 17

Ex 5.3, 18 Important

Ex 5.3, 19 Important

Ex 5.3, 20 Important

Chapter 5 Class 10 Arithmetic Progressions

Serial order wise

Last updated at April 16, 2024 by Teachoo

Ex 5.3, 10 Show that a1, a2 … , an , … form an AP where an is defined as below (i) an = 3 + 4n .Also find the sum of first 15 terms an = 3 + 4n Taking n = 1 a1 = 3 + 4 × 1 a = 3 + 4 a = 7 Taking n = 2 a2 = 3 + 4 × 2 a2 = 3 + 8 a2 = 11 Taking n = 3 a3 = 3 + 4 × 3 a3 = 3 + 12 a3 = 15 Now we have to find sum of first 15 terms i.e. S15 We know that Sn = 𝑛/2 (2𝑎+(𝑛−1)𝑑) Here, n = 15, a = 7, d = 4 Putting the values S15 = 15/2 (2 ×7+(15−1)×4) = 15/2 (14 + 14 × 4) = 15/2 (14+46) = 15/2×70 = 15 × 35 = 525 Hence series is 7, 11, 15, ….. Since difference is same, it is an AP Common difference = d = 4