F.6 Geometric sequence

31.
(a) (i)
A1B = 3a ´[2/(1 + 2)] cm = 2a cm
B1B = 3a ´[1/(1 + 2)] cm = a cm

By Pythagorean theorem:
A1B1² = A1B² + B1B²
A1B1² = [(2a)² + a²] cm²
A1B1² = 5a² cm²
A1B1 = (√5)a cm

(a) (ii)
A2B1 = (√5)a ´ [2/(1 + 2)] cm = 2(√5)a/3 cm
B2B1 = (√5)a ´ [1/(1 + 2)] cm = (√5)a/3 cm

By Pythagorean theorem:
A2B2² = A2B1² + B2B1²
A2B2² = [(2(√5)a/3)² + ((√5)a/3)²] cm²
A2B2² = 25a²/9 cm²
A2B2 = 5a/3 cm

(a) (iii)
A3B2 = (5a/3) ´ [2/(1 + 2)] cm = 10a/9 cm
B3B2 = (5a/3) ´ [1/(1 + 2)] cm = 5a/9 cm

By Pythagorean theorem:
A3B3² = A3B2² + B3B2²
A3B3² = [10a/9)² + (5a/9)²]cm²
A3B3² = 125a²/81 cm²
A3B3 = 5(√5)a/9 cm


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(b)
Area of square A1B1­C1D1 = [(√5)a]² cm² = 5a² cm²
Area of square A2B2C2D2 = (5a/3)² cm² = 5²a²/9 cm²
Area of square A3B3C3D3 = [5(√5)a/9]² cm² = 5³a²/9² cm²

It can be deduced that :
Area of square AnBnCnDn = 5^n a^2 / 9^(n - 1) cm²


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(c) (i)
CD = 3a cm = 6 cm
Hence, a = 2

Area of square A5B5C5D5
= (5)^5 (2)^2 / 9^4 cm²
= 1.9 cm² (to 1 sig. fig.)

(c) (ii)
Areas (in cm²) of squares A1B1C1D1,A2B2C2D2, A3B3C3D3...... form a geometric sequence.
T(1) = a = 5(2)² = 20
r = [5^n a^2 / 9^(n - 1)] / 5^(n - 1) a^2 / 9^(n - 2) = 5/9

The total area of the squares A1B1C1D1,A2B2C2D2, ......, A5B5C5D5
= a(1 - r^5) / (1 - r)
= 20[1 - (5/9)^5] / [1 - (5/9)] cm²
= 43 cm² (to2 sig. fig.)

(c) (iii)
The total area of the squares A1B1C1D1,A2B2C2D2, A3B3C3D3
= a / (1 - r)
= 20 / [1 - (5/9)] cm²
= 45 cm²

2011-10-15 22:02:38 補充：
The first line of (c) (iii) should be :
The total area of the squares A1B1C1D1,A2B2C2D2, A3B3C3D3 ......