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what is the recursive rule for this geometric sequence?

3, 3/2, 3/4, 3,8,...

choices -
a1= 1/2;an = 3/2 x an - 1
a1= 3/2;an = 1/2 x an -1
a1= 3; an = 1/2 x an -1
a1= 1/2;an = 3x an -1

please explain your answer, if you can

Respuesta :

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

Answer:

C.[tex]a_1=3,a_n=\frac{1}{2}a_{n-1},n\geq 2[/tex]

Step-by-step explanation:

We are given that

[tex]3,\frac{3}{2},\frac{3}{4},\frac{3}{8}[/tex],..

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

[tex]a_1=3[/tex]

[tex]a_2=\frac{3}{2}[/tex]

[tex]a_2=3\times \frac{1}{2}=\frac{1}{2}\times a_1[/tex]

[tex]a_3=\frac{3}{4}[/tex]

[tex]a_3=\frac{1}{2}\times \frac{3}{2}[/tex]

[tex]a_3=\frac{1}{2}\times a_2[/tex]

[tex]a_4=\frac{3}{8}[/tex]

[tex]a_4=\frac{1}{2}\times \frac{3}{4}[/tex]

[tex]a_4=\frac{1}{2}\times a_3[/tex]

Therefore, the recursive rule

[tex]a_1=3,a_n=\frac{1}{2}a_{n-1},n\geq 2[/tex]

Option C is true

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