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 You are here

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

Chapter 5 Class 10 Arithmetic Progressions

Serial order wise

Last updated at April 16, 2024 by Teachoo

Ex 5.3, 3 In an AP (vi) Given a = 2, d = 8, Sn = 90, find n and an. Given a = 2, d = 8, Sn = 90 We can use formula Sn = 𝑛/2 (2𝑎+(𝑛−1)𝑑) Putting a = 2, d = 8, Sn = 90 90 = 𝑛/2 (2 × 2+(𝑛−1) × 8) 90 = 𝑛/2 (4+8𝑛−8) 90 × 2 =𝑛(4+8𝑛−8) 180 = n (8n – 4) 180 = 8n2 – 4n –180 + 8n2 – 4n = 0 8n2 – 4n – 180 = 0 8n2 – 4n – 180 = 0 4 (2n2 – n – 45) = 0 2n2 – n – 45 = 0 2n2 – 10n + 9n – 45 = 0 2n(n – 5) + 9(n – 5) = 0 (2n + 9) (n − 5) = 0 2n + 9 = 0 2n = –9 n = (−𝟗)/𝟐 n – 5 = 0 n = 5 Therefore, n = 5 & n = (−9)/2 But n cannot be in fraction, So, n = 5 We need to find an, i.e. a5 We know that an = a + (n – 1)d Putting a = 2, n = 5, d = 8 a5 = 2 + (5 – 1) 8 a5 = 2 + (4)8 a5 = 2 + 32 a5 = 34 Therefore, n = 5 & a5 = 34