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Ex 5.3, 3 (v) - Chapter 5 Class 10 Arithmetic Progressions
Last updated at May 29, 2023 by Teachoo
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
Transcript
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
