# Give the formula for the following sequence: 4, 12, 36, ...

Give the formula for the following sequence: 4, 12, 36, ...

Since 4 x 3 = 12, and 12 x 3 = 36, you can determine that this is a geometric sequence in which the common ratio is 3.

Geometric sequences are written in the form: ```y=a•r^(n-1)` where y = term, a = 1st term, r = common ratio.

Therefore this equation would be:  `y=4xx3^(n-1)` ``

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There can be a lot of ways to figure these types of problems out.  One thing that is going on, one can tell, is that we are multiplying each number by 3 to get the next number (3*4=12, 12*3 = 36).  But, that doesn't tell us a formula.

Given that this is multiplication, we see that this is a geometric sequence.  So, the formula would be of the form y = a*b^x.  We are looking for numbers for a and b.  The way we do that is first consider how the numbers are arranged, as in:

n number  y

1          4

2          12

3          36

Where "n number" represents "the first number", the second number", etc.  And, y is the actual number.  So, then, we can have coordinate points (1,4), (2,12), and (3,36).

Plugging any 2 of these into the general formula would give us something like:

4 = a*b^1          and          12 = a*b^2

The first one simplifies to:

4 = a*b       or a = 4/b

Substituting that into the second equation, we get:

12 = (4/b)*b^2 = 4*(b^2)/b = 4*b

12 = 4b

So, b = 3.

Subbing this into the first equation we made, 4 = a*b:

4 = a*3

a = 4/3

So, the formula for this sequence is:

y = (4/3)*3^x          or, we can use n for the number of the term

y = (4/3)*3^n

So, then, if we want the first number, n = 1

y = 4/3 * 3^1 = 4/3 * 3 = 4

For the second number, n = 2

y = (4/3) * 3^2 = (4/3) * 9 = 12

And, so on.

==============================================Or, it can also be written as y = a*r^(n-1), where a is the first term and r is the what we are multiplying each number by, the common multiplier.  We stated earlier that is 3.  And, the first term is 4.  So, another way to write the formula is:

y = a*r^(n-1)

y = 4*3^(n-1)

And, we would get the same numbers as before.

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