



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 Deleted for CBSE Board 2022 Exams
Ex 5.3, 16 Important Deleted for CBSE Board 2022 Exams
Ex 5.3, 17 Deleted for CBSE Board 2022 Exams
Ex 5.3, 18 Important Deleted for CBSE Board 2022 Exams
Ex 5.3, 19 Important Deleted for CBSE Board 2022 Exams
Ex 5.3, 20 Important Deleted for CBSE Board 2022 Exams
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