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Ex 5.3, 3 (i) - 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) You are here
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
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 (i) Given a = 5, d = 3, an = 50, find n and Sn. Given a = 5 , d = 3 , an = 50 We know that an = a + (n – 1) d Putting values 50 = 5 + (n – 1) ×3 50 = 5 + 3n – 3 50 = 2 + 3n 50 – 2 = 3n 48 = 3n 48/3=𝑛 n = 16 Now we need to find Sn Sn = 𝒏/𝟐(𝟐𝒂+(𝒏−𝟏)𝒅) Putting n = 16, a = 5, d = 3 = 16/2 (2 × 5+(16−1) × 3) = 8 (10+15 × 3) = 8(10+45) = 8 ×55 = 440 So, answer is n = 7 and a = –8
