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Ex 5.3, 3 - Chapter 5 Class 10 Arithmetic Progressions - Part 10

Ex 5.3, 3 - Chapter 5 Class 10 Arithmetic Progressions - Part 11
Ex 5.3, 3 - Chapter 5 Class 10 Arithmetic Progressions - Part 12

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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 (πŸπ’‚+(π’βˆ’πŸ)𝒅) Putting Sn = S9 = 75 , n = 9 , d = 5 75 = 9/2 (2π‘Ž+(9βˆ’1) Γ— 5) 75 = 9/2 (2π‘Ž+8 Γ— 5) 75 = 9/2(2π‘Ž+40) (75 Γ— 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 (βˆ’210)/18=π‘Ž (βˆ’35)/3=π‘Ž a = (βˆ’πŸ‘πŸ“)/πŸ‘ Now, we need to find a9 We know that an = a + (n – 1) d Putting a = ( 35)/3, d = 5 & n = 9 a9 = (βˆ’35)/3 + (9 βˆ’ 1) Γ— 5 a9 = (βˆ’35)/3 + 8 Γ— 5 a9 = (βˆ’35)/3 + 40 a9 = (βˆ’35 + 40 Γ— 3)/3 a9 = (βˆ’35 + 120)/3 a9 = πŸ–πŸ“/πŸ‘

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