Question

what kind of sequence is formed by the areas of the center triangles in the sequence of figures shown?
arithmetic, with common difference $\frac{1}{4}$
arithmetic, with common difference 4
geometric, with common ratio $\frac{1}{4}$
neither arithmetic or geometric

Explanation:

Step1: Analyze the area - change pattern

Let the area of the largest triangle in the first figure be \(A_1\). In the second figure, the center - triangle is formed by dividing the largest triangle into 4 equal - sized sub - triangles, so if the area of the center - triangle in the second figure is \(A_2\), then \(A_2=\frac{1}{4}A_1\). In the third figure, the new center - triangle is formed from the center - triangle of the second figure in the same way. If the area of the center - triangle in the third figure is \(A_3\), then \(A_3 = \frac{1}{4}A_2\).

Step2: Recall the definitions of arithmetic and geometric sequences

An arithmetic sequence has a common difference \(d\) such that \(a_{n + 1}-a_n=d\). A geometric sequence has a common ratio \(r\) such that \(\frac{a_{n + 1}}{a_n}=r\). Here, since \(\frac{A_2}{A_1}=\frac{1}{4}\), \(\frac{A_3}{A_2}=\frac{1}{4}\), the sequence of the areas of the center - triangles has a common ratio \(r = \frac{1}{4}\).

Answer:

Geometric, with common ratio \(\frac{1}{4}\)