


Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
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
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
Last updated at May 29, 2023 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