Question 1196057
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Without more info, there could be infinitely many possible answers. See my previous post <a href="https://www.algebra.com/algebra/homework/Sequences-and-series/Sequences-and-series.faq.question.1195913.html">here</a> for more details.


If we assume the sequence is geometric, then,
a = 32 = first term
r = -1/2 = common ratio
The idea is to multiply each term by -1/2


which leads to,
nth term = a*( r )^(n-1)
nth term = <font color=red>32*( -1/2 )^(n-1)</font>
Again this all hinges on the assumption we have a geometric sequence. 


Let's check the answer with something like n = 3
a_n = 32*(-1/2)^(n-1)
a_3 = 32*(-1/2)^(3-1)
a_3 = 32*(-1/2)^(2)
a_3 = 32*(1/4)
a_3 = 8
We have 8 as the third term, which indeed matches with the given sequence. I'll let you check the other values.
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