Question

Enter the explicit rule for the geometric sequence.

9,6,4,83,…

Answer 1

Answer:

a_n = 2^(n - 1) 3^(3 - n)

Step-by-step explanation:

9,6,4,8/3,…

a1 = 3^2

a2 = 3 * 2

a3 = 2^2

As we can see, the 3 ^x is decreasing and the 2^ y is increasing

We need to play with the exponent in terms of n

Lets look at the exponent for the base of 2

a1 = 3^2  2^0

a2 = 3^1   2^1

a3 =    3^ 0 2^2

an =   3^      2^(n-1)

I picked n-1  because that is where it starts  0

 n = 1   (1-1) =0

  n=2   (2-1) =1

  n=3  (3-1) =2

Now we need to figure out the exponent for the 3 base

I will pick (3-n)

n =1  (3-1) =2

n =2  (3-2) =1

n=3    (3-3) =0

Answer 2

We want to find the explicit rule for the given geometric sequence. We will get the explicit rule:

$A_n = (2/3)^{n - 1}*9$

Remember that in a geometric sequence, each term is given by a constant multiplying the previous term.

For the second and third terms we have:

4 = k*6

k = 4/6 = 0.6... and so on.

So we know that:

k = 4/6 = 2/3.

Now that we know this value, we can write the recursive rule as:

$A_n = (2/3)*A_{n-1}$

This means that the n-th term is (2/3) times the previous one.

The explicit rule will be just:

$A_n = k^{n-1}*A_1 = (2/3)^{n-1}*9$

This just gives us the value of the n-th term in the sequence as a function of k, n, and the first term on the sequence, A₁.

If you want to learn more about geometric sequences, you can read:

https://brainly.com/question/9300199