81, 27, 9, 3,... Find the common ratio of the given sequence, and write an exponential function which represents the sequence. Use n = 1, 2, 3, ... A) 3; f(n) = 81n-1 B) 3; f(n) = 81(3)n-1 Eliminate C) 1 3 ; f(n) = 81(3)n-1 D) 1 3 ; f(n) = 81( 1 3 )n-1

The options are not written clearly, and they are difficult to choose from. I will however, find the common ratio, and write an exponential function which represents the given sequence.

Step-by-step explanation:

The sequence is: 81, 27, 9, 3, ...

This is a geometric progression that has a clear pattern.

Let S1 = 81, S2 = 27, S3 = 9, and so on.

To find the common ratio of this sequence - is to find the ratio S2 to S1, or S3 to S2, or S4 to S3, and so on. If there is common ratio, then

S2/S1 = S3/S2 = S4/S3 = ...

Now let us check.

S2/S1 = 27/81 = 1/3

S3/S2 = 9/27 = 1/3

S4/S3 = 3/9 = 1/3

This is enough to conclude that the common ratio is 1/3

We now want to find an exponential function that represents the sequence. That is, we want to find an expression which defines each term of this sequence, the nth term of the sequence.

Note that the nth term of a geometric progression is given as

S_n = ar^(n - 1)

Where a is the first term

r is the common ratio.

For this sequence, a = 81, and r = 1/3.

Let us obtain the nth term at once.

S_n = 81(1/3)^(n - 1)

This is what you are looking for.

You can test it but putting n = 1, 2, 3,... you will have your sequence back.

1. 3

2. 4

3. 1

hope this ! have a great day.

-wfz

maybe i can