Need an answer as soon as possible. asking for a friend. use the sequence: 1, –2, 4, –8, . . a. find the common ratio in the sequence. b. write a rule for the sequence. c. find the next three terms

Part A

The common ratio is -2



Explanation: 

To get the common ratio r, we divide any term by the previous one

One example:
r = common ratio
r = (second term)/(first term)
r = (-2)/(1)
r = -2

Another example:
r = common ratio
r = (third term)/(second term)
r = (4)/(-2)
r = -2
and we get the same common ratio every time

Side Note: each term is multiplied by -2 to get the next term


Part B

The rule for the sequence is 
a(n) = (-2)^(n-1)
where n starts at n = 1



Explanation:

Recall that any geometric sequence has the nth term
a(n) = a*(r)^(n-1)
where the 'a' on the right side is the first term and r is the common ratio

The first term given to use is a = 1 and the common ratio found in part A above was r = -2
So,
a(n) = a*(r)^(n-1)
a(n) = 1*(-2)^(n-1)
a(n) = (-2)^(n-1)


Part C

The next three terms are 16, -32, 64



Explanation:

We can simply multiply each previous term by -2 to get the next term. Do this three times to generate the next three terms

-8*(-2) = 16
16*(-2) = -32
-32*(-2) = 64

showing that the next three terms are 16, -32, and 64

An alternative is to use the formula found in part B

Plug in n = 5 to find the fifth term
a(n) = (-2)^(n-1)
a(5) = (-2)^(5-1)
a(5) = (-2)^(4)
a(5) = 16 which matches with what we got earlier

Then plug in n = 6
a(n) = (-2)^(n-1)
a(6) = (-2)^(6-1)
a(6) = (-2)^(5)
a(6) = -32 which matches with what we got earlier

Then plug in n = 7
a(n) = (-2)^(n-1)
a(7) = (-2)^(7-1)
a(7) = (-2)^(6)
a(7) = 64 which matches with what we got earlier

while the second method takes a bit more work, its handy for when you want to find terms beyond the given sequence (eg: the 28th term)

(-14+3/2b) - (1+8/2b) = -5 -5/2b(-10+b) + (7b-5) = 8b-15(7/2b-3) - (8+6b) = -5/2b-11(5+2b) + (2b+3/2) = 4b+ 13/2