To determine the type of sequence represented by the table, we first need to analyze the relationship between the successive [tex]\( y \)[/tex]-values.
Given the table:
[tex]\[
\begin{array}{|c|c|c|c|c|}
\hline
x & 1 & 2 & 3 & 4 \\
\hline
y & 5 & 3 & 1.8 & 1.08 \\
\hline
\end{array}
\][/tex]
Step 1: Identify the successive [tex]\( y \)[/tex]-values.
- The [tex]\( y \)[/tex]-values are: 5, 3, 1.8, 1.08
Step 2: Check for an arithmetic sequence.
- In an arithmetic sequence, the difference between successive [tex]\( y \)[/tex]-values is constant.
- Calculate the differences:
- [tex]\( y_2 - y_1 = 3 - 5 = -2 \)[/tex]
- [tex]\( y_3 - y_2 = 1.8 - 3 = -1.2 \)[/tex]
- [tex]\( y_4 - y_3 = 1.08 - 1.8 = -0.72 \)[/tex]
- Since the differences are not constant ([tex]\(-2, -1.2, -0.72\)[/tex]), the sequence is not arithmetic.
Step 3: Check for a geometric sequence.
- In a geometric sequence, the ratio between successive [tex]\( y \)[/tex]-values is constant.
- Calculate the ratios:
- [tex]\( \frac{y_2}{y_1} = \frac{3}{5} = 0.6 \)[/tex]
- [tex]\( \frac{y_3}{y_2} = \frac{1.8}{3} = 0.6 \)[/tex]
- [tex]\( \frac{y_4}{y_3} = \frac{1.08}{1.8} = 0.6 \)[/tex]
- Since the ratios are all equal to 0.6, the sequence is geometric with a common ratio of 0.6.
Thus, the correct answer is:
B. The table represents a geometric sequence because the successive [tex]\( y \)[/tex]-values have a common ratio of 0.6.
