1. A sequence of the form $\left\{t_{1}, t_{2}, \ldots, t_{n}\right\}$ is called geometric if $t_{1}=a, t_{2}=a r, t_{3}=$ $a r^{2}, \ldots, t_{n}=a r^{n-1}$. For example, $\{1,2,4,8,16\}$ and $\{1,-3,9,-27\}$ are both geometric sequences. In all three questions below, suppose $\left\{t_{1}, t_{2}, t_{3}, t_{4}, t_{5}\right\}$ is a geometric sequence of real numbers.
(a) If $t_{1}=3$ and $t_{2}=6$, determine the value of $t_{5}$.
(b) If $t_{2}=2$ and $t_{4}=8$, determine all possible values of $t_{5}$.
(c) If $t_{1}=32$ and $t_{5}=2$, determine all possible values of $t_{4}$.
Solution:
(a) $t_{1}=3=a$ and $t_{2}=a r=6$, so $r=6 / 3=2$.
This gives $t_{5}=3 \times 2^{4}=48$.
(b) $t_{2}=2=a r$ and $t_{4}=8=a r^{3}$, Dividing the two equations gives $r^{2}=4$, so $r= \pm 2$.
When $r=2$ we have $a=1$, so $t_{5}=2^{4}=16$. .
When $r=-2$ we have $a=-1$, so $t_{5}=-1 \times 2^{4}=-16$.
(c) We have $t_{1}=32=a$ and $t_{5}=2=a r^{4}$. This gives $a=32$, and $r^{4}=\frac{1}{16}$.
When $r^{4}=\frac{1}{16}$ we get $r^{2}=\frac{1}{4}$ or $r^{2}=\frac{-1}{4}$.
When $r^{2}=\frac{-1}{4}, r$ is not a real number, so this is not a valid sequence.
When $r^{2}=\frac{1}{4}$ we get $r= \pm \frac{1}{2}$.
This gives $t_{4}=32 \times \frac{1}{8}=4$ and $t_{4}=32 \times \frac{-1}{8}=-4$.