Question

Which exponential function can be used to show the geometric sequence 40, 160, 640, . . . ?(1 point)
Responses

an=40(4)n−1
𝑎
𝑛
=
40
(
4
)
𝑛
−
1
a subscript n baseline equals 40 left parenthesis 4 right parenthesis superscript n minus 1 baseline

an=40+4(n−1)
𝑎
𝑛
=
40
+
4
(
𝑛
−
1
)
a subscript n baseline equals 40 plus 4 left parenthesis n minus 1 right parenthesis

an=4(40)n−1
𝑎
𝑛
=
4
(
40
)
𝑛
−
1
a subscript n baseline equals 4 left parenthesis 40 right parenthesis superscript n minus 1 baseline

an=40(120)n−1

Answer 1

To determine the exponential function that represents the geometric sequence 40, 160, 640, ..., we first need to identify the first term and the common ratio.

1. The first term (when n=1) is \( a_1 = 40 \).
2. The second term (when n=2) is \( a_2 = 160 \).
3. The third term (when n=3) is \( a_3 = 640 \).

To find the common ratio (\( r \)):
- From \( 40 \) to \( 160 \): \( 160 / 40 = 4 \)
- From \( 160 \) to \( 640 \): \( 640 / 160 = 4 \)

Thus, the common ratio is \( r = 4 \).

The general formula for a geometric sequence can be written as:
\[ a_n = a_1 \cdot r^{n-1} \]

Where:
- \( a_1 \) is the first term (40 in this case),
- \( r \) is the common ratio (4 in this case),
- \( n \) is the term number.

Using the values we identified:
\[ a_n = 40 \cdot 4^{n-1} \]

Thus, the correct function to represent the geometric sequence is:
\[ a_n = 40(4)^{n-1} \]

So, the response is:
**an = 40(4)^(n−1)**.