F.6 math arithmetic sequence~~

There are 60 similar figures. The perimeter of the 1st figure is 5 cm.The perimeter of each succeeding figure is 2 cm longer than that of theprevious one. It is known that the area of the 1st figure is 2cm².
(a) Find the areas of the 2nd, 4th and 6th figures each.
(b) Do the areas of the 2nd, 4th, 6th,...,60th figures from an arithmeticsequence? Explain briefly.

Solution :
(a)
T(n) cm : The perimeter of the nth figure, where
T(1) = 5
Common difference = 2

T(2) = 5 + (2 - 1)x2 = 7
T(4) = 5 + (4 - 1)x2 = 11
T(6) = 5 + (6 - 1)x2 = 15

ratio of areas of two similar figures it equal to the ratio of the squares ofperimeters.

(Area of the 1st figure) : (Area of the nth figure)
= (Perimeter of the 1st figure)² : (Perimeter of the nthfigure)²

(2 cm²) : (Area of the 2nd figure) = (5 cm)² : (7 cm)²
Area of the 2nd figure = 2 x 7² / 5² = 3.92 cm²

(2 cm²) : (Area of the 4th figure) = (5 cm)² : (11 cm)²
Area of the 4th figure = 2 x 11² / 5² = 9.68 cm²

(2 cm²) : (Area of the 6th figure) = (5 cm)² : (15 cm)²
Area of the 6th figure = 2 x 15² / 5² = 18 cm²

(b)
(Area of the 4th figure) - (Area of the 2nd figure)
= (9.68 - 3.92)²
= 5.76 cm²

(Area of the 6th figure) - (Area of the 4th figure)
= (18 - 9.68)²
= 8.32 cm²

The differences are not equal,
thus the areas of the 2nd, 4th, 6th,...,60th figures do not from anarithmetic sequence.