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 You are here

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 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 (iv) Given a3 = 15, S10 = 125, find d and a10. For a3 = 15 We know that an = a + (n β 1) d Putting an = a3 = 15 & n = 3 15 = a + 2d 15 β 2d = a a = 15 β 2d For S10 = 125 We know that Sn = π/2(π+(πβ1)π) Putting Sn = S10 = 125, n = 10 S10 = 10/2 (2π+(10β1)π) 125 = 5 (2a + 9d) 125/5=2π+9π 25 = 2a + 9d 25 β 9d = 2a a = (ππ β ππ )/π From (1) and (2) 15 β 2d = (ππβππ )/π 30 β 4d = 25 β 9d 5 = β 5d 5/(β5)=π d = β1 Putting value of d in (1) a = 15 β 2d a = 15 β 2 Γ(β1) a = 15 + 2 a = 17 Now we need to find a10 an = a + (n β 1) d Putting n = 10 , a = 17 & d = β1 a10 = 17 + (10 β 1) Γ (β1) a10 = 17 + 9 Γ (β1) a10 = 17 β 9 a10 = 8