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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) You are here

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

Chapter 5 Class 10 Arithmetic Progressions

Serial order wise

Last updated at May 29, 2023 by Teachoo

Ex 5.3, 3 In an AP (v) Given d = 5, S9 = 75, find a and a9. Given d = 5 , S9 = 75 We use formula Sn = π/π (ππ+(πβπ)π ) Putting Sn = S9 = 75 , n = 9 , d = 5 75 = 9/2 (2π+(9β1) Γ 5) 75 = 9/2 (2π+8 Γ 5) 75 = 9/2(2π+40) (75 Γ 2)/9 = 2a + 40 150/9 = 2a + 40 150 = 9 (2a + 40) 150 = 9 (2a) + 9(40) 150 = 18a + 360 150 β 360 = 18a β210 = 18a (β210)/18=π (β35)/3=π a = (βππ)/π Now, we need to find a9 We know that an = a + (n β 1) d Putting a = ( 35)/3, d = 5 & n = 9 a9 = (β35)/3 + (9 β 1) Γ 5 a9 = (β35)/3 + 8 Γ 5 a9 = (β35)/3 + 40 a9 = (β35 + 40 Γ 3)/3 a9 = (β35 + 120)/3 a9 = ππ/π