
Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
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) You are here
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
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
Last updated at May 29, 2023 by Teachoo
Ex 5.3, 3 In an AP (ii) Given a = 7, a13 = 35, find d and S13. Given a = 7, a13 = 35 We need to find d We know that an = a + (n โ 1) d Putting a = 7, n = 13 and an = 35 35 = 7 + (13 โ 1) ร ๐ 35 = 7 + 12d 35 โ 7 = 12d 28 = 12 d 28/12=๐ 7/3=๐ d = ๐/๐ Now we need to find S13 We can use formula Sn = ๐/๐ (๐+๐) Putting n = 13, a = 7, ๐ = a13 = 35 = 13/2(7+35) = 13/2 ร 42 = 13 ร 21 = 273