Last updated at Dec. 13, 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
Ex 5.3
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
About the Author
Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo