Question

A geometric sequence starts with the values an=7,21,63,189 . . .
. What is the exponential equation for this sequence?(1 point)
y=
(
)x

Answer 1

To find the exponential equation for the given geometric sequence \( a_n = 7, 21, 63, 189, \ldots \), we first identify the common ratio of the sequence.

1. **Calculate the common ratio:**
\[
r = \frac{a_2}{a_1} = \frac{21}{7} = 3
\]
\[
r = \frac{a_3}{a_2} = \frac{63}{21} = 3
\]
\[
r = \frac{a_4}{a_3} = \frac{189}{63} = 3
\]
The common ratio \( r \) is 3.

2. **Find the general term of the geometric sequence:**
The general form of a geometric sequence can be expressed as:
\[
a_n = a_1 \cdot r^{(n-1)}
\]
where \( a_1 \) is the first term, and \( r \) is the common ratio.

Here, \( a_1 = 7 \) and \( r = 3 \), so:
\[
a_n = 7 \cdot 3^{(n-1)}
\]

3. **Express it in terms of \( y \) and \( x \):**
If we let \( n = x \), then:
\[
y = 7 \cdot 3^{(x-1)}
\]
This can also be rewritten as:
\[
y = 7 \cdot 3^x \cdot 3^{-1}
\]
\[
y = \frac{7}{3} \cdot 3^x
\]

However, the standard form can simply be kept as:
\[
y = 7 \cdot 3^{(x-1)}
\]

Thus, the exponential equation for this sequence is:
\[
y = 7 \cdot 3^{(x-1)}
\]