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 given geometric sequence $a_n=7,21,63,189,\dots$, 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 2

y = BLANK (BLANK)x

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 $B=3$.