急! F6 Geometric Sequences 02q7

14(a) The area of C1=π x 1^2=π cm^2
The area of C2=π x (1x2)^2=4π cm^2
The area of C3=π x (1x2x2)^2=16π cm^2

(b) The area of C2/The area of C1=4π/π=4
The area of C3/The area of C2=16π/4π=4
The areas of the circles form a geometric sequence.
The total area of the circles (to C7)
=π(4^7-1)/(4-1). (First term=π, common ratio=4)
=π(16383)/3
=5461π cm^2

19(a) R be the ratio of the length of the pieces.
T6=64R^5=2
R^5=1/32
R=1/2
The length of the original string
=64(1-(1/2)^6)/(1-1/2). (first term=64, common ratio=1/2)
=126 cm
(b) Let r be the common ratio.
T1+T2+T3=126
18+18r+18r^2=126
18(1+r+r^2)=126
1+r+r^2=7
r^2+r-6=0
(r-2)(r+3)=0
r=2 or r=-3 (rejected)
The length of th longest string
=18 x 2^2
=72 cm