Ex 5.3, 3 (i) - Chapter 5 Class 10 Arithmetic Progressions
Last updated at Dec. 13, 2024 by Teachoo
Last updated at Dec. 13, 2024 by Teachoo
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
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) 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)
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Ex 5.3, 10 (i)
Ex 5.3, 10 (ii) Important
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Ex 5.3, 14 Important
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Ex 5.3, 16 Important
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