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) You are here

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 (vii) Given a = 8, an = 62, Sn = 210, find n and d. Given a = 8, an = 62, Sn = 210, Since there are n terms, 𝑙 = an = 62 We use the formula Sn = 𝒏/𝟐 (𝒂+𝒍) Putting a = 8, Sn = 210, 𝑙 = an = 62 210 = 𝑛/2 (8+62) 210 × 2=𝑛 × (70) 420 = n × 70 420/70=𝑛 6 = n n = 6 Now we need to find d We can use formula an = a + (n – 1) d Putting an = 62, a = 8, n = 6 62 = 8 + (6 – 1) × 𝑑 62 = 8 + 5d 62 – 8 = 5d 54 = 5d d = 𝟓𝟒/𝟓