Question

What is the nth term of the geometric sequence 4, 8, 16, 32, ... ?

Answer 1

A. 2(4)^N-1
B. 4(2)^N-1
C. 2(N)^4
D. 4(N)^2

Answer 2

formula is

(first term)*(common ratio)^(N-1)

what is the first term?

Answer 3

\[a _{n}\]

Answer 4

not quite, take a look at your sequence

what is the first number in the list?

Answer 5

ditto

Answer 6

A. \[A. a_{n}=2(4)^{n-1} B. a_{n}=4(2)^{n-1} C.a_{n}=2(n)^{4} D. a_{n}=4(n)^{2}\]2(4)^N-1
B. 4(2)^N-1
C. 2(N)^4
D. 4(N)^2

Answer 7

would you mind answering my question, please?

your sequence is 4, 8, 16, 32, ... , right?
what is the first number in the sequence?

Answer 8

idk

Answer 9

which number comes first in the list 4, 8, 16, 32, ... ?

Answer 10

4

Answer 11

right! that means our first term is 4

our sequence is 4, 8, 16, 32, ...

each consecutive number gets multiplied by 2 to get to the next number, right? our common ratio is 2

so, first term = 4, and common ratio = 2

now, let's look at the formula (first term)*(common ratio)^(N-1)

can you tell me which one is the answer now? :)

Answer 12

A

Answer 13

not quite, check carefully



(first term)*(common ratio)^(N-1)
first term is 4
common ratio is 2

Answer 14

B

Answer 15

much better, good job

Answer 16

What is the tenth term of the geometric sequence that has a common ratio of 1/3 and 36 as its fifth term?
A. 4/7
B. 27/4
C. 4/81
D. 1/36