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)

Ex 5.3, 3 (viii) Important You are here

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 (viii) Given an = 4, d = 2, Sn = −14, find n and a. Given an = 4, d = 2, Sn = –14 Since there are n terms, 𝑙 = an = 4 We use the formula Sn = 𝒏/𝟐 (𝒂+𝒍) Putting Sn = −14, 𝑙 = an = 4 –14 = 𝑛/2 (𝑎+4) –14 × 2 =𝑛(𝑎+4) –28 = n (a + 4) (−28)/(𝑎 + 4)=𝑛 n = (−𝟐𝟖)/(𝒂 + 𝟒) Also we know that an = a + (n – 1) d Putting an = 4 , d = 2 4 = a + (n – 1) × 2 4 = a + 2n – 2 4 + 2 = a + 2n 6 = a + 2n Putting n = (− 𝟐𝟖)/(𝒂 + 𝟒) 6 = a + 2((− 28)/(𝑎 + 4)) 6 = a − 56/(𝑎 + 4) 6 = (𝑎(𝑎 + 4) − 56)/(𝑎 + 4) 6 = (𝑎2 + 4𝑎 − 56)/(𝑎 + 4) 6 (a + 4) = a2 + 4a – 56 6a + 24 = a2 4a – 56 0 = a2 + 4a – 56 – 6a – 24 0 = a2 – 2a – 80 a2 – 2a – 80 = 0 a2 – 10a + 8a– 80 = 0 a (a – 10) + 8 (a – 8) = 0 (a – 10) (a + 8) = 0 So, a = 10 or a = –8 Taking a = 10 n = (− 28)/(𝑎 + 4) n = (− 28)/(10 + 4) n = (− 28)/14 n = –2 Since n is number of terms, it cannot be negative So, n = –2 is not possible ∴ a = 10 is not possible Taking a = – 8 n = (− 28)/(𝑎 + 4) n = (− 28)/(− 8 + 4) n = (− 28)/( − 4) n = 7 So, n = 7