Find the 31st term of an arithmetic sequence where the eighth term is 45 and the fourth term is 65.

Let a (n) denote the nth term in the sequence. Then

a (5) = a (4) + d

a (6) = a (5) + d = a (4) + 2d

a (7) = a (6) + d = a (4) + 3d

a (8) = a (7) + d = a (4) + 4d

where d is the fixed difference between consecutive terms in the sequence.

We have a (4) = 65 and a (8) = 45, so we solve for d:

45 = 65 + 4d = => d = - 5

so, given some term in the sequence a (n), the next term is a (n + 1) = a (n) - 5.

Continuing the pattern above, we would wind up getting

a (31) = a (4) + 27d

(In case the pattern isn't clear, watch the coefficient of d; then 1 + 4 = 5, 2 + 4 = 6, 3 + 4 = 7, and so on.)

Then the 31st term in the sequence must be

a (31) = 65 + 31 (-5) = - 90