Ex 5.3, 10 (i) - Chapter 5 Class 10 Arithmetic Progressions
Last updated at April 16, 2024 by Teachoo
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
Transcript
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
