Question

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

Answer 1

In a geometric sequence, each term after the first is found by multiplying the previous term by a fixed, non-zero number called the common ratio.

Let's look at the provided sequence:
- \( a_1 = 7 \)
- \( a_2 = 21 \)
- \( a_3 = 63 \)
- \( a_4 = 189 \)

First, we can find the common ratio \( r \):
- To go from \( 7 \) to \( 21 \): \( r = \frac{21}{7} = 3 \)
- To go from \( 21 \) to \( 63 \): \( r = \frac{63}{21} = 3 \)
- To go from \( 63 \) to \( 189 \): \( r = \frac{189}{63} = 3 \)

Thus, the common ratio \( r \) is \( 3 \).

In a geometric sequence, the \( n \)-th term can be expressed as:
\[
a_n = a_1 \cdot r^{(n-1)}
\]

Substituting in our values:
- \( a_1 = 7 \)
- \( r = 3 \)

So the formula for the \( n \)-th term becomes:
\[
a_n = 7 \cdot 3^{(n-1)}
\]

To express this in the form of an exponential equation \( y = ()x \), we can relate it to \( y \) and \( x \) by letting \( x = n \):
\[
y = 7 \cdot 3^{(x-1)}
\]

This can also be simplified:
\[
y = \frac{7}{3} \cdot 3^x
\]

If you want a simpler version focusing just on \( n \) as a variable, you can write:
\[
y = 7 \cdot 3^{x-1}
\]

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