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

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Answer 1 · 2016-08-13T15:08:47

#a_32 = -171#

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#