3,4,6,15,9,48,12....

The odd numbered terms are just

3,  6,   9,   12,...  which is easy, just the multiples of 3.

The even numbered term are

  4,  15,  48

So it amounts to finding the next term of 4,15,48,...

You can easily find a quadratic polynomial that has those 
first three terms:

11n^2-22n+15

11(1)^2-22(1)+15 = 11-22+15 = 4
11(2)^2-22(2)+15 = 44-44+15 = 15
11(3)^2-22(3)+15 = 99-66+15 = 48

So the next term is:

11(4)^2-22(4)+15 = 176-88+15 = 103

So 103 is certainly a possible answer.  But there are many other
possibilities.  

Does your teacher know that there are an infinite number of different
sequences that have all those given first seven terms?  If so, how does 
your teacher expect you to pick out the very one he has in mind?

Edwin