Geometric Sequence 1條

(a) Radius of P1 = 10 (given)
Let Center of circle be O
OB1 = 10
OA2 = radius of circle P2.
For equilateral triangle, OA2/OB1 = sin 30 = 1/2, so radius of P2 = OA2 = OB1/2 = 5.
Similarly, radius of P3 = 5/2, radius of P4 = 5/4 and so on.
So sum of circumference = 2p(10) + 2p(5) + 2p(5/2) + 2p(5/4) + .....
= 2p(10 + 5 + 5/2 + 5/4 + ......)
= 2p[ 10/(1 - 1/2)] = 2p(20) = 40p. ( p = pi).
(b)
A2B1/OB1 = cos 30 = (sqrt 3)/2
So perimeter of equilateral triangle = 6 x A2B1 = (6)[ sqrt(3)/2](10)
= 30 sqrt 3.
Perimeter of 2nd triangle = 3 x A2B1 = 15 sqrt 3
So sum of perimeter = 30 sqrt 3 + 15 sqrt 3 + (15 sqrt 3)/2 + ......
sqrt 3 ( 30 + 15 + 15/2 + 15/4 + .....) = sqrt 3 [ 30/(1 - 1/2)] = 60 sqrt 3.