How do you find #a_12# given 8, 3, -2, ...?

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Answer 1 · 2016-12-25T19:50:55

#a_12=-47#

Find #a_12# given #8,3,-2...#

Note that each term is 5 less than the previous term. This implies that the sequence is arithmetic and of the form

#a_n=a_1+(n-1)d#

where #a_1# is the first term and #d# is the common difference between the terms.

In this example #a_1=8# and #d=-5#

The "formula" for this sequence is then

#a_n=8+(n-1)(-5)#
#a_n=8-5n+5#
#a_n=-5n+13#

To find #a_12# (the 12th term), plug in #12# for #n#.

#a_12=-5(12)+13=-60+13=-47#