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

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