Question

What is the recursive rule for this geometric sequence?

2, 1/2, 1/8. 1/32,  ...
Enter your answers in the boxes.
an= __ ⋅an−1
a1= __

Thank you for your time!

Answer 1

`a_(n)=(1)/(4)a_(n-1)\\a_(1)=2`

Explanation:

`a_(n1)=2,\ a_(2)=(1)/(2),\ a_(3)=(1)/(8),\ a_(4)=(1)/(32)\\\\(a_(2))/(a_(1))=((1)/(2))/(2)=(1)/(4)\\\\(a_(3))/(a_(2))=(((1)/(8)))/(((1)/(2)))=(1)/(4)\\\\(a_(4))/(a_(3))=(((1)/(32)))/(((1)/(8)))=(1)/(4)`

So

this is a geometric series with first term
`=2`

and constant rate
`(1)/(4)`

`n^(th)` term is given by
`a_(n)=2((1)/(4))^(n-1)`

`(n-1)^(th)\ term=2((1)/(4))^(n-1-1)=2((1)/(4))^(n-2)\\\\(a_n)/(a_(n-1))=(2((1)/(4))^(n-1))/(2((1)/(4))^(n-2))}=((1)/(4)((1)/(4))^(n-2))/(((1)/(4))^(n-2))}=(1)/(4)\\\\(a_(n))/(a_(n-1))=(1)/(4)\\\\a_n=(1)/(4)a_(n-1)`