Only 4 points have been given: 3, 6, 7, -2, so there may be many ways to derive the equation which calculates the n-th term.
Here's one method. First note that this is not a linear progression. Let's assume a cubic progression:
T(n)=an3+bn2+cn+d where a, b, c, d are constant coefficients. Let the sequence start with n=0, so T(0)=3=d and:
T(1)=6=a+b+c+d, or a+b+c=3 because we know that d=3.
T(2)=7=8a+4b+2c+d, 8a+4b+2c=4.
T(3)=-2=27a+9b+3c+3, 27a+9b+3c=-5.
T(2)-2T(1)=6a+2b=-2⇒3a+b=-1.
T(3)-3T(1)=24a+6b=-14⇒12a+3b=-7.
T(3)-3T(1)-3(T(2)-2T(1))=3a=-4, a=-4/3⇒b=-1-(-4)=3⇒c=3-(-4/3+3)=4/3.
Therefore, T(n)=-4n3/3+3n2+4n/3+3, which can be written T(n)=⅓(-4n3+9n2+4n+9).