Question 983064
The n-th term of an arithmetic progression can be written a_n = a_1 + (n-1)d,
where a_1 = the first term and d = the common difference.
In this case a_1 = -7, and d = 3
For the last term in the sequence we can write 101 = -7 + 3(n-1)
Solve for n:
101 = 3n - 10
n = 111/3 = 37
So there are 37 terms in the sequence.