Question

What is the recursive rule for this geometric sequence? 3, 3/2, 3/4, 3/8

Answer 1

I think that would be  an  = an-1 *  1/2    where an = nth term, and an-1 = previous term. 

Answer 2

Answer:

$a_{n}=\frac{a_{n-1}}{2}$

Step-by-step explanation:

Let's look a the sequence:

$3, \frac{3}{2} ,\frac{3}{4} ,\frac{3}{8}$

we can see that

$\frac{3}{2}=3*\frac{1}{2}$

and

$\frac{3}{4} =\frac{3}{2}*\frac{1}{2}$

and

$\frac{3}{8}=\frac{3}{4} *\frac{1}{2}$

so each next number is equal to the previous number multiplied by $\frac{1}{2}$ or, in other other words, is the previous number divided by 2.

So the recursive formula must be:

$a_{n}=\frac{a_{n-1}}{2}$

where $a_{n}$ is a number in the sequence, and $a_{n-1}$ is the immediate previous number.