## College Algebra 7th Edition

geometric sequence $a_5=4\sqrt{2}$
We are given: $\sqrt{2}, 2, 2\sqrt{2}, 4$, ... We can see that this is a geometric sequence by dividing the terms: $\frac{a_2}{a_1}=\frac{2}{\sqrt{2}}=\sqrt{2}$ $\frac{a_3}{a_2}=\frac{2\sqrt{2}}{2}=\sqrt{2}$ $\frac{a_4}{a_3}=\frac{4}{2\sqrt{2}}=\sqrt{2}$ Thus $r=\sqrt{2}$. So the 5th term is: $a_{5}=a_{4}*r=4*\sqrt{2}=4\sqrt{2}$