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International Baccalaureatte | Further Maths HL

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##### Nelson Loo

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Anyone out there knows hw to solve qus 1 and 4, thxs

Date Posted: 11 months ago
Views: 71
J
11 months ago
Sn = a(rⁿ -1)/(r - 1)

S2n = a(r²ⁿ - 1)/(r - 1)

S2n/Sn = a(r²ⁿ -1)/(r - 1) ÷ a(rⁿ - 1)/(r - 1) = K

a(r²ⁿ -1)/(r - 1) x (r - 1) / (a(rⁿ - 1)) = K

(r²ⁿ - 1) / (rⁿ - 1) = K

(rⁿ + 1)(rⁿ - 1) / (rⁿ - 1) = K

rⁿ + 1 = K (proved)

S = a / (1 - r)

2S = a² / (1 - r²)

25/8 = a + ar = a(1 + r) -----①

2S / S = a² / (1 - r²) ÷ a / (1 - r)

2 = a² / (1 - r²) x (1 - r) / a

2 = a/(1 + r)
a = 2(1 + r) ----②

Sub ② into ①,

25/8 = 2(1 + r)(1 + r) = 2r² + 4r + 2

2r² + 4r - 9/8 = 0

r² + 2r - 9/16 = 0

(r - ¼)(r + 9/4) = 0

r = ¼ or r = -9/4 (N.A, ∣ r ∣ < 1 for series to be convergent)

so r = ¼.
so a = 2(1 + ¼) = 5/2

S = a / (1 - r) = 5/2 / (1 - ¼)

S = 10/3 = 3⅓

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Date Posted: 11 months ago 