Question

A geometric sequence starts with the values: an = 7, 21, 63, 189.... What is the exponential equation for this sequence?

Answer 1

y = BLANK (BLANK)x

Answer 2

To find the exponential equation for the given geometric sequence \( a_n = 7, 21, 63, 189, \ldots \), we start by identifying the first term and the common ratio.

1. **Identify the first term \( a_1 \):**
- The first term \( a_1 = 7 \).

2. **Identify the common ratio \( r \):**
- To find the common ratio, we divide the second term by the first term:
\[
r = \frac{a_2}{a_1} = \frac{21}{7} = 3
\]
- We can verify this by checking the ratio between subsequent terms:
\[
r = \frac{a_3}{a_2} = \frac{63}{21} = 3
\]
\[
r = \frac{a_4}{a_3} = \frac{189}{63} = 3
\]
- The common ratio \( r \) is consistently \( 3 \).

3. **Write the general formula for the \( n \)-th term of a geometric sequence:**
The \( n \)-th term of a geometric sequence can be given by the formula:
\[
a_n = a_1 \cdot r^{n-1}
\]
Substituting the values we found:
- \( a_1 = 7 \)
- \( r = 3 \)

Thus, the equation becomes:
\[
a_n = 7 \cdot 3^{n-1}
\]

This is the exponential equation for the given geometric sequence.

Answer 3

From your prompt, it seems you want the exponential equation in the format \( y = A \cdot B^x \).

Given the sequence we discussed, the equation we derived is:

\[
a_n = 7 \cdot 3^{n-1}
\]

To write it in the format \( y = A \cdot B^x \), we can make a substitution for \( n \):

Let \( x = n - 1 \). Then \( n = x + 1 \), and we can rewrite the equation:

\[
a_n = 7 \cdot 3^{(x + 1) - 1} = 7 \cdot 3^x
\]

Thus, the equation in the desired format is:

\[
y = 7 \cdot 3^x
\]

So we can fill in your blanks as:

\( y = 7 \cdot (3^x) \)

Where \( A = 7 \) and \( B = 3 \).