In a geometric sequence with five terms $t_{1}, t_{2}, t_{3}, t_{4}, t_{5}$, each term is positive and $t_{1}>t_{2}$. If the sum of the first two terms is $\frac{15}{2}$ and the sum of the squares of the first two terms is $\frac{153}{4}$, what is the value of $t_{5}$ ?

(A geometric sequence is a sequence in which each term after the first is obtained from the previous term by multiplying it by a non-zero constant. For example, 3, 6, 12 is a geometric sequence with three terms.)
Suppose that the given sequence has first term $t_{1}=a$ and common ratio $r$.

Thus, $t_{2}=a r$.

Since $t_{1}+t_{2}=\frac{15}{2}$, then $a+a r=\frac{15}{2}$ or $a(1+r)=\frac{15}{2}$, from which we obtain $a=\frac{15}{2(1+r)}$. (Note that $r$ cannot equal -1 .)

Since $\left(t_{1}\right)^{2}+\left(t_{2}\right)^{2}=\frac{153}{4}$, then $a^{2}+a^{2} r^{2}=\frac{153}{4}$, or $a^{2}\left(1+r^{2}\right)=\frac{153}{4}$.

Substituting for $a$, we obtain

$$
\begin{aligned}
\left(\frac{15}{2(1+r)}\right)^{2}\left(1+r^{2}\right) & =\frac{153}{4} \\
\frac{225\left(1+r^{2}\right)}{4(1+r)^{2}} & =\frac{153}{4} \\
\frac{25\left(1+r^{2}\right)}{(1+r)^{2}} & =17 \\
25+25 r^{2} & =17\left(1+2 r+r^{2}\right) \\
8 r^{2}-34 r+8 & =0 \\
4 r^{2}-17 r+4 & =0 \\
(r-4)(4 r-1) & =0
\end{aligned}
$$

Thus, $r=4$ or $r=\frac{1}{4}$.

Since $t_{1}>t_{2}$, then $r<1$, so $r=\frac{1}{4}$.

Since $a=\frac{15}{2(1+r)}$, then $a=\frac{15}{2\left(\frac{5}{4}\right)}=6$.

Finally, $t_{5}=a r^{4}=6\left(\frac{1}{4}\right)^{4}=\frac{6}{4^{4}}=\frac{6}{256}=\frac{3}{128}$.

ANSWER: $\frac{3}{128}$