Ex 5.3
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
Last updated at April 16, 2024 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