Question

What is the recursive rule for this geometric sequence?

Answer 1

Answer:

C.$a_1=3,a_n=\frac{1}{2}a_{n-1},n\geq 2$

Step-by-step explanation:

We are given that

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

We have to find the recursive formula rule for this geometric sequence.

$a_1=3$

$a_2=\frac{3}{2}$

$a_2=3\times \frac{1}{2}=\frac{1}{2}\times a_1$

$a_3=\frac{3}{4}$

$a_3=\frac{1}{2}\times \frac{3}{2}$

$a_3=\frac{1}{2}\times a_2$

$a_4=\frac{3}{8}$

$a_4=\frac{1}{2}\times \frac{3}{4}$

$a_4=\frac{1}{2}\times a_3$

Therefore, the recursive rule

$a_1=3,a_n=\frac{1}{2}a_{n-1},n\geq 2$

Option C is true

Answer 2

The objective here is to find $r$ (so called common ratio):
$r=a_{n}/a_{n-1}=a_{2}/a_{1}= \frac{3}{2} : 3 = \frac{3}{2} * \frac{1}{3} = \frac{1}{2}$
So assuming the first element of sequence is 3 (as you mentioned) we now can define the recursive rule for this geometric sequence:
$a_{1}=3; a_{n}=\frac{1}{2}a_{n-1}$