With these sequence-type problems, I usually find it helpful to make lists and look for patterns so it will be easier to fill in the blanks.
First, identify clearly what they're asking for: the # of new members in week 12. (They call this y, but that turns out to be unimportant.)
Then, start with the information they've given you:
There are 5 new members in week 1. Each new member brings x new members the following week.
Let's start with a sample number to see how this sequence works. Let's say each new member brings 2 new members the following week. How many new members will there be in week 2?
5 new members from week 1 x 2 new members each = 10 new members for week 2
How many new members will there be in week 3?
10 new members from week 2 x 2 new members each = 20 new members for week 3
Aha! We're just multiplying by 2 each time! Therefore, week 4 will be
5*2*2*2 = 5 x 2^3 = 40 new members for week 4
We start to see a pattern...
# new members in week n will be 5*2^(n-1)
However, we just picked 2 as a test number. Let's generalize that finding by saying the following:
# new members in week n will be 5*x^(n-1)
Therefore, the number of new members in week 12 will be 5*x^(12-1) = 5*x^(11)
We don't know what x is, but at least we know the format of the calculation. We then scan the answer choices looking for a 5 times some number to the 11th power.
A) This number is going to be way too small. This is the twelfth root of 5.
B) If we try to manipulate this to get it into the form we want, we end up with 5*3^11*5^10. That's not what we want.
C) Again, if we try to get this into our target form, we have 5*3^12*5^11. Again, not what we want
D) If we take a 5 from this we have 5*3^11*5^11. Again, not quite what we want. BUT this one we can manipulate further to get what we want! Remember this property:
(a^x)*(b^x) = (ab)^x
Therefore (3^11)*(5^11) = 15^11.
Eureka! We have 5*15^11, which means Answer D is a possibility, and x=15.
E) We could get 5*12^12, which is again, not what we're looking for.
The correct answer is D.
