The first term of two geometric sequences is equal, the difference between the second, third, and fourth terms is $5, -\frac{5}{4}$, and $\frac{35}{16}$, respectively. Which are these two geometric sequences?
The common first term of the two geometric sequences is $a$, the common ratio of the first sequence is $p$, and that of the second is $q$.

The difference of the second terms

$$
a p-a q=a(p-q)=5
$$

the difference of the third terms

$$
a p^{2}-a q^{2}=a\left(p^{2}-q^{2}\right)=-\frac{5}{4}
$$

the difference of the fourth terms

$$
a p^{3}-a q^{3}=a\left(p^{3}-q^{3}\right)=\frac{35}{16}
$$

Dividing (2) by (1)

$$
p+q=-\frac{1}{4}
$$

Dividing (3) by (1)

$$
p^{2}+p q+q^{2}=\frac{7}{16}
$$

Subtracting (5) from the square of (4)

$$
p q=\frac{1}{16}-\frac{7}{16}=-\frac{6}{16}=-\frac{3}{8}
$$

Based on (4) and (6), $p$ and $q$ are the roots of the equation

$$
x^{2}+\frac{1}{4} x-\frac{3}{8}=0
$$

Thus,

$$
p_{1}=q_{2}=\frac{1}{2}, \quad \text { and } \quad p_{2}=q_{1}=-\frac{3}{4}
$$

From (1)

$$
\begin{gathered}
a=\frac{5}{p-q}=\frac{5}{ \pm \frac{5}{4}}= \pm 4 \\
a_{1}=4 \text { when } p>q, \text { so } p_{1}=\frac{1}{2}, q_{1}=-\frac{3}{4}
\end{gathered}
$$

and thus the two sequences we are looking for are:

$$
\begin{gathered}
4,2,1, \frac{1}{2}, \ldots, \text { or } 4,-3, \frac{9}{4},-\frac{27}{16}, \ldots \\
a_{2}=-4 \text { when } p<q, \text { so } p_{2}=-\frac{3}{4}, q_{2}=\frac{1}{2},
\end{gathered}
$$

and thus the second solution gives us the sequences

$$
-4,3,-\frac{9}{4}, \frac{27}{6}, \ldots \quad \text { or }-4,-2,-1,-\frac{1}{2}, \ldots
$$