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 Aug. 3, 2021 by Teachoo

Ex 5.3, 3 In an AP (viii) Given an = 4, d = 2, Sn = β14, find n and a. Given an = 4, d = 2, Sn = β14 Since there are n terms, π = an = 4 We use the formula Sn = π/π (π+π) Putting Sn = β14, π = an = 4 β14 = π/2 (π+4) β14 Γ 2 =π(π+4) β28 = n (a + 4) (β28)/(π + 4)=π n = (βππ)/(π + π) Also we know that an = a + (n β 1) d Putting an = 4 , d = 2 4 = a + (n β 1) Γ 2 4 = a + 2n β 2 4 + 2 = a + 2n 6 = a + 2n Putting n = (β ππ)/(π + π) 6 = a + 2((β 28)/(π + 4)) 6 = a β 56/(π + 4) 6 = (π(π + 4) β 56)/(π + 4) 6 = (π2 + 4π β 56)/(π + 4) 6 (a + 4) = a2 + 4a β 56 6a + 24 = a2 4a β 56 0 = a2 + 4a β 56 β 6a β 24 0 = a2 β 2a β 80 a2 β 2a β 80 = 0 a2 β 10a + 8aβ 80 = 0 a (a β 10) + 8 (a β 8) = 0 (a β 10) (a + 8) = 0 So, a = 10 or a = β8 Taking a = 10 n = (β 28)/(π + 4) n = (β 28)/(10 + 4) n = (β 28)/14 n = β2 Since n is number of terms, it cannot be negative So, n = β2 is not possible β΄ a = 10 is not possible Taking a = β 8 n = (β 28)/(π + 4) n = (β 28)/(β 8 + 4) n = (β 28)/( β 4) n = 7 So, n = 7