The 12th term of a sequence is 41 and the 15th term is 140. what is the nth term?

An= mth term.
an=a₁+(n-1)*d

a₁₂=41
a₁₅=140

a₁₂=41
41=a₁+(12-1)*d
41=a₁+11d 
a₁+11d=41    (1)

a₁₅=140
140=a₁+(15-1)*d
140=a₁+14d 
a₁+14d=140    (2)

With the equiations (1) and (2)  build a system of equations
a₁+11d=41
a₁+14d=140

we solve it.
-(a₁+11d=41)
  a₁+14d=140

       3d=99  ⇒d=99/3=33

a₁+11d=41
a₁+(11*33)=41
a₁+363=41
a₁=41-363=-322

an=a₁+(n-1)*d
an=-322+(n-1)*33
an=-322+33n-33
an=-355+33n

an=-355+33n

To check:
a₁₂=-355+33*12=-355+396=41
a₁₅=-355+33*15=-355+495=140.