What is the 32nd term of the arithmetic sequence where a_1 = 15 and a_13= –57?

1 Answer
George C.
Aug 13, 2016

a_32 = -171

Explanation:

The general formula for a term of an arithmetic sequence is:

a_n = a+d(n-1)

where a is the initial term and d the common difference.

So:

-72 = -57-15 = a_13 - a_1 = (a+12d)-(a+0d) = 12d

Hence:

d = -6

a = a_1 = 15

So:

a_32 = a+31d = 15 - 6*31 = -171

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