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 Deleted for CBSE Board 2022 Exams

Ex 5.3, 16 Important Deleted for CBSE Board 2022 Exams

Ex 5.3, 17 Deleted for CBSE Board 2022 Exams

Ex 5.3, 18 Important Deleted for CBSE Board 2022 Exams

Ex 5.3, 19 Important Deleted for CBSE Board 2022 Exams

Ex 5.3, 20 Important Deleted for CBSE Board 2022 Exams

Chapter 5 Class 10 Arithmetic Progressions (Term 2)

Serial order wise

Last updated at Aug. 3, 2021 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 = π/2 (2π+(πβ1)π) Putting a = 2, d = 8, Sn = 90 90 = π/2 (2 Γ 2+(πβ1) Γ 8) 90 = π/2 (4+8πβ8) 90 Γ 2 =π(4+8πβ8) 180 = n (8n β 4) 180 = 8n2 β 4n β180 + 8n2 β 4n = 0 8n2 β 4n β 180 = 0 8n2 β 4n β 180 = 0 4 (2n2 β n β 45) = 0 2n2 β n β 45 = 0 2n2 β 10n + 9n β 45 = 0 2n(n β 5) + 9(n β 5) = 0 (2n + 9) (n β 5) = 0 2n + 9 = 0 2n = β9 n = (βπ)/π n β 5 = 0 n = 5 Therefore, n = 5 & n = (β9)/2 But n cannot be in fraction, So, n = 5 We need to find an, i.e. a5 We know that an = a + (n β 1)d Putting a = 2, n = 5, d = 8 a5 = 2 + (5 β 1) 8 a5 = 2 + (4)8 a5 = 2 + 32 a5 = 34 Therefore, n = 5 & a5 = 34