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Ex 5.3, 3 - Chapter 5 Class 10 Arithmetic Progressions - Part 13

Ex 5.3, 3 - Chapter 5 Class 10 Arithmetic Progressions - Part 14
Ex 5.3, 3 - Chapter 5 Class 10 Arithmetic Progressions - Part 15
Ex 5.3, 3 - Chapter 5 Class 10 Arithmetic Progressions - Part 16

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Ex 5.3, 3 In an AP (vi) Given a = 2, d = 8, Sn = 90, find n and an. Given a = 2, d = 8, Sn = 90 We can use formula Sn = 𝑛/2 (2π‘Ž+(π‘›βˆ’1)𝑑) Putting a = 2, d = 8, Sn = 90 90 = 𝑛/2 (2 Γ— 2+(π‘›βˆ’1) Γ— 8) 90 = 𝑛/2 (4+8π‘›βˆ’8) 90 Γ— 2 =𝑛(4+8π‘›βˆ’8) 180 = n (8n – 4) 180 = 8n2 – 4n –180 + 8n2 – 4n = 0 8n2 – 4n – 180 = 0 8n2 – 4n – 180 = 0 4 (2n2 – n – 45) = 0 2n2 – n – 45 = 0 2n2 – 10n + 9n – 45 = 0 2n(n – 5) + 9(n – 5) = 0 (2n + 9) (n βˆ’ 5) = 0 2n + 9 = 0 2n = –9 n = (βˆ’πŸ—)/𝟐 n – 5 = 0 n = 5 Therefore, n = 5 & n = (βˆ’9)/2 But n cannot be in fraction, So, n = 5 We need to find an, i.e. a5 We know that an = a + (n – 1)d Putting a = 2, n = 5, d = 8 a5 = 2 + (5 – 1) 8 a5 = 2 + (4)8 a5 = 2 + 32 a5 = 34 Therefore, n = 5 & a5 = 34

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