Ex 5.3, 3 (v) - Chapter 5 Class 10 Arithmetic Progressions
Last updated at April 16, 2024 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 = 𝒏/𝟐 (𝟐𝒂+(𝒏−𝟏)𝒅) Putting Sn = S9 = 75 , n = 9 , d = 5 75 = 9/2 (2𝑎+(9−1) × 5) 75 = 9/2 (2𝑎+8 × 5) 75 = 9/2(2𝑎+40) (75 × 2)/9 = 2a + 40 150/9 = 2a + 40 150 = 9 (2a + 40) 150 = 9 (2a) + 9(40) 150 = 18a + 360 150 – 360 = 18a –210 = 18a (−210)/18=𝑎 (−35)/3=𝑎 a = (−𝟑𝟓)/𝟑 Now, we need to find a9 We know that an = a + (n – 1) d Putting a = ( 35)/3, d = 5 & n = 9 a9 = (−35)/3 + (9 − 1) × 5 a9 = (−35)/3 + 8 × 5 a9 = (−35)/3 + 40 a9 = (−35 + 40 × 3)/3 a9 = (−35 + 120)/3 a9 = 𝟖𝟓/𝟑
