Abe has a bag of marbles and wants to divide them into sets. The first set will have 2 marbles; the second set, 5; the third set, 8; the fourth set, 11; and so on. How many sets will he have if he starts with 40 marbles?

The number of marbles in each set is an arithmetic sequence. The sum of any arithmetic sequence is that average of the first and last terms times the number of terms ... mathematically this is:

s (n) = (2an+dn^2-dn) / 2, a=initial term, n=term number, d=common difference

In this case the first term is 2 and since each term is 3 more than the previous term, the common difference is 3, so now we can say:

s (n) = (2*2n+3n^2-3n) / 2

s (n) = (3n^2+n) / 2 and we are told that there is a total of 40 so

(3n^2+n) / 2=40

3n^2+n=80

3n^2+n-80=0

3n^2-15n+16n-80=0

3n (n-5) + 16 (n-5) = 0

(3n+16) (n-5) = 0, since n>0

n=5

So Abe will make five sets of marbles ...

(The sets contain 2,5,8,11, and 14 marbles, which sum to 40 marbles total)

He will have 5 sets because 2 plus 5 plus 8 plus 11 plus 14 is 40