急 F6 Geometric Sequences 02q14

32.
(a)(i)
The first term, a = 8
The common ratio, r = 4/2 = 8/4 = 2

His answer
= S9
= a [1 - r⁹] / [1 - r]
= 8 [1 - (1/2)⁹] / [1 - (1/2)]
= 15.96875

(a)(ii)
The correct answer
= S∞
= a / [1 - r]
= 8 / [1 - (1/2)]
= 16

(a)(iii)
The percentage error
= [(15.96875 - 16)/16] x 100%
= -0.20% (2 sig. fig.)

(b)
16 - Sk < 0.002
16 - {8 [1 - (1/2)ᵏ] / [1 - (1/2]} < 0.002
-16 [1 - (1/2)ᵏ] < -15.998
1 - (1/2)ᵏ > (-15.998)/(-16)
(1/2)ᵏ < 1 - (15.998/16)
log(1/2)ᵏ < log[1 - (15.998/16)]
k log(1/2) < log[1 - (15.998/16)]
k > log[1 - (15.998/16)] / log(1/2)
k > 12.97

The minimum value of k = 13


31.
(a)(i)
By Pythagorean theorem, in ΔA1BB1 :
A1B1 = √[(2a)² + (a)²) cm = √5a cm

(a)(ii)
By Pythagorean theorem, in ΔA2B1B2 :
A2B2 = √[(2√5a/3)² + (√5a/3)²] = 5a/3 cm

(a)(iii)
By Pythagorean theorem, in ΔA3B2B3 :
A2B2 = √[(10a/9)² + (5a/9)²] = 5√5a/9 cm

(b)
Area of A1B1C1D1 = (√5a)² cm² = 5a² cm²
Area of A2B2C2D2 = (5a/3)² cm² = 5a² x (5/9) cm²
Area of A3B3C3D3 = (5√5a/9)² cm² = 5a² x (5/9)² cm²

Hence, area of AnBnCnDn = 5a² x (5/9)ⁿ⁻¹ cm²

(c)(i)
AB = CD = 6 cm
3a = 6
a = 2

<Area of An­BnCnDn> (in cm²)is a geometric sequence.
The first term, T1 = 5a² = 5(2)² = 20
The common ratio, r = 5/9

Area of A5B5C5D5
= 20 x (5/9)⁴ cm²
= 12500/6561 cm²

(c)(ii)
The required total area
= a [1 - r⁵] / [1 - r]
= 20 x [1 - (5/9)⁵] / [1 - (5/9)]
=279620/6561

(c)(iii)
The required total area
= a / [1 - r]
= 20 / [1 - (5/9)]
=45