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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

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 May 29, 2023 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 – 1) d Putting an = a3 = 15 & n = 3 15 = a + 2d 15 – 2d = a a = 15 – 2d For S10 = 125 We know that Sn = 𝑛/2(𝑎+(𝑛−1)𝑑) Putting Sn = S10 = 125, n = 10 S10 = 10/2 (2𝑎+(10−1)𝑑) 125 = 5 (2a + 9d) 125/5=2𝑎+9𝑑 25 = 2a + 9d 25 – 9d = 2a a = (𝟐𝟓 − 𝟗𝒅)/𝟐 From (1) and (2) 15 – 2d = (𝟐𝟑−𝟗𝒅)/𝟐 30 – 4d = 25 – 9d 5 = – 5d 5/(−5)=𝑑 d = –1 Putting value of d in (1) a = 15 – 2d a = 15 – 2 ×(−1) a = 15 + 2 a = 17 Now we need to find a10 an = a + (n – 1) d Putting n = 10 , a = 17 & d = –1 a10 = 17 + (10 – 1) × (−1) a10 = 17 + 9 × (−1) a10 = 17 – 9 a10 = 8