急! F6 Geometric Sequences 02q5

9.
Let Tn cm be the perimeter of the nth rectangle.

T1 = 2 x (20 + 12) = 64
T2 = 2 x [(1/2)x20 + (1/2)x12] = (1/2)T1
T3 = 2 x [(1/2)²x20 + (1/2)²x12]= (1/2)²T1
......

Hence, T1, T2, T3, ...... is a geometricsequence.
The first term, a = T1 = 64
The common ratio, r = (1/2)

T6
= ar⁵
= 64 x (1/2)⁵
= 2

The perimeter of the 6th rectangle = 2cm


=====
10.
By mid-point theorem :
A2B2 = (1/2)A1B1, A3B­3= (1/2)A2B2, ...... A­nBn = (1/2)An-1Bn-1

Let Tn­ cm be the perimeter of the nth triangle (ΔAnBnCn).

T1 = 3 x 10 = 30
T2 = 3 x [(1/2) x (10)] = (1/2)T1
T3 = 3 x [(1/2)² x 10] = (1/2)²T1
......
Tn = 3 x [(1/2)ⁿ⁻¹ x 10] = (1/2)ⁿ⁻¹T1

Hence, T1, T2, T3, ...... is a geometricsequence.
The first term, a = T1 = 30
The common ratio, r = (1/2)

Sum to infinite terms
= a / [1 - r]
= 30 / [1 - (1/2)]
= 60

The total length of the perimeters = 60cm