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

Answer 1

To find the exponential equation for the geometric sequence given by the terms \( a_n = 7, 21, 63, 189, \ldots \), we first need to determine the common ratio.

The common ratio \( r \) can be found by dividing any term by its previous term:

\[ r = \frac{a_2}{a_1} = \frac{21}{7} = 3 \] \[ r = \frac{a_3}{a_2} = \frac{63}{21} = 3 \] \[ r = \frac{a_4}{a_3} = \frac{189}{63} = 3 \]

So the common ratio \( r \) is 3.

The first term \( a_1 \) is 7.

The general formula for a geometric sequence is given by:

\[ a_n = a_1 \cdot r^{n-1} \]

Substituting in the values we have:

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

This is the exponential equation for the geometric sequence.

Answer 2

That is incorrect it needs to be in the form of y=____(____)^x

Answer 3

I apologize for the misunderstanding. To rewrite the equation in the form \( y = A (B)^x \), we can adjust the equation we found for the geometric sequence.

Given the general formula we derived:

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

We can express this in the form \( y = A (B)^x \). In this case:

• Let \( y = a_n \),
• \( A = 7 \),
• \( B = 3 \), and
• we need to adjust for the exponent.

Since \( n \) starts from 1, we can replace \( n \) with \( x + 1 \) (so \( n = x + 1 \) implies \( x = n - 1 \)) to fit the form you specified. Therefore:

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

So the final equation in the form \( y = A (B)^x \) is:

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

Thank you for your patience!