I think that this is a combination problem. From the given, the 8 students are taken 3 at a time. This can be solved through using the formula of combination which is C(n,r) = n!/(n-r)!r!. In this case, n is 8 while r is 3. Hence, upon substitution of the values, we have
C(8,3) = 8!/(8-3)!3!
C(8,3) = 56
There are 56 3-person teams that can be formed from the 8 students.
An=Asub1+d(n-1)
Asub5= -5+½(4)
=-5+7
=2
-1/2 + 2 = 1.5
because if u add 2 to a minus half it goes 0 and then 1.5 (positive half)
The best way to answer this would be making a graph
For 1 sign it costs 54.50 and he makes 15
For 2 signs it costs 59 and he makes 30
For 3 signs it costs 63.50 and he makes 45
For 4 signs it costs 68 and he makes 60
For 5 signs it costs 72.50 and he makes 75
This means that after 5 signs Larry will be making profit
