Question

What is the 15th term of the sequence 4, -8, 16, -32, 64, ...?

Answer 1

`\bf 4~~,~~\stackrel{-2\cdot 4}{-8}~~,~~\stackrel{-2\cdot -8}{16}~~,~~\stackrel{-2\cdot 16}{-32}~~,~~\stackrel{-2\cdot -32}{64}~~...\impliedby r=-2 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ n^(th)\textit{ term of a geometric sequence} \\\\ a_n=a_1\cdot r^(n-1)\qquad \begin{cases} n=n^(th)\ term\\ a_1=\textit{first term's value}\\ r=\textit{common ratio}\\[-0.5em] \hrulefill\\ a_1=4\\ r=-2\\ n=15 \end{cases} \\\\\\ a_(15)=4(-2)^(15-1)\implies a_(15)=4(-2)^(14)\implies a_(15)=65536`

Answer 2

I believe the answer is -32,768