Answer:
[tex]a_{14} = 10496[/tex]
Step-by-step explanation:
A geometric sequence is of the form:
[tex]ar^0, ar^1, ar^2, ar^3, \cdot \cdot \cdot[/tex]
So the given values can be represent as:
[tex]1. \quad ar^5 = \frac{41}{256}\\2. \quad ar^{10} = -164[/tex]
So you have a system of equations with two unknows [tex]a[/tex] and [tex]r[/tex]. Then you can find one unknow in terms of the other.
[tex]a = \frac{-164}{r^{10}}[/tex]
Now replace [tex]a[/tex] in the first equation
[tex]\frac{-164}{r^{10}}r^5 = \frac{41}{256}\\\frac{-164}{r^{5}} = \frac{41}{256}\\r^5 = \frac{256 \cdot -164}{41} = -\frac{41984}{41}\\r = \sqrt[5]{-\frac{41984}{41}}\\[/tex]
You don't have to worry about the negative number inside the root, because is a odd root then you can find odd negative numbers that give me the desire result.
[tex]r = -\frac{4\sqrt[5]{41} }{\sqrt[5]{41}} \\r = -4[/tex]
For find [tex]a[/tex] only replace [tex]r[/tex] in whatever of the two equations.
[tex]-a4^5 = \frac{41}{256}\\-a1024 = \frac{41}{256}\\a = -\frac{41}{262144}[/tex]
After find [tex]a[/tex] and [tex]r[/tex] use the next formula for find the [tex]14^{th}[/tex] term.
[tex]ar^{13} = x\\-\frac{41}{262144} \cdot -4^{13} = x\\\\10496 = x[/tex]
