How do you write a rule for the nth term of the arithmetic sequence and then find #a_22# given #45, 31, 17, 3,...#?

1 Answer
Sep 22, 2017

Answer:

#a_n = 45 - (14) xx n#
#a_22 = -263#

Explanation:

First, we figure out the sequence. Then we generalize it.
This one shows a decrease of 14 (-14) between every value in the series.

So, if we start with #a_0 = 45# we see that #a_1 = a_0 - 14#.
The next one would be #a_2 = a_1 - 14#. Substituting our value for #a_1# from the previous difference we see that:

#a_2 = (a_0 - 14) - 14# or #a_2 = a_0 - (14) xx 2#

Recognizing that now our multiplier is the same as our sequence identifier (2) we can generalize the expression to:

#a_n = a_0 - (14) xx n#; or #a_n = 45 - (14) xx n#

NOW putting in our values for #a_22# we can obtain:
#a_22 = a_0 - (14) xx 22# ; #a_22 = 45 - 308 = -263#