Answer:
1.C
2.A
Step-by-step explanation:
We know:
and
These expressions have an 
term in common. We can combine them if we can find forms of the expressions in terms of :
Now, because 
(by the reflexive property), we can write:
This is our answer, A.
Part A
Answer: 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
Answer:
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
Answer: 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)
