The sum of the first few terms of a geometric sequence is 11, the sum of their squares is 341, and the sum of their cubes is 3641. Determine the terms of the sequence.
Let the first element of the sequence be $a$, the ratio $q$, and the number of terms $n$. $q$ cannot be 1, because then $an = 11$, and from $a^{2} n=341$ we would have $a=31, n=\frac{11}{31}$, which is not an integer. Thus, we can write the following equations:

$$
a \cdot \frac{q^{n}-1}{q-1}=11
$$

$$
a^{2} \cdot \frac{q^{2 n}-1}{q^{2}-1}=341
$$

$$
a^{3} \cdot \frac{q^{3 n}-1}{q^{3}-1}=3641
$$

Dividing the second equation by the square of the first, we get:

$$
\frac{\left(q^{n}+1\right)(q-1)}{\left(q^{n}-1\right)(q+1)}=\frac{31}{11}
$$

From this, $q^{n}=\frac{21 q+10}{10 q+21}$. Substituting this into (1) yields $a=10 q+21$.

Substituting the expressions for $q^{n}$ and $a$ into (3), and using the identity

$$
x^{3}-y^{3}=(x-y)\left(x^{2}+x y+y^{2}\right)
$$

we get:

$$
\begin{gathered}
(10 q+21)^{3} \frac{\left(\frac{21 q+10}{10 q+21}\right)^{3}-1}{q^{3}-1}=3641 \\
\frac{(21 q+10)^{3}-(10 q+21)^{3}}{q^{3}-1}=3641 \\
\frac{(11 q-11)\left((21 q+10)^{2}+(21 q+10)(10 q+21)+(10 q+21)^{2}\right)}{(q-1)\left(q^{2}+q+1\right)}=11 \cdot 331
\end{gathered}
$$

We can simplify by $q-1 \neq 0$, and then divide by 11, and after rearranging, we arrive at the equation $2 q^{2}+5 q+2=0$. From this, the possible values for $q$ are $q=-2$, and $q=-\frac{1}{2}$. (This is not surprising, since if a sequence is a solution with a certain $q$ as the ratio, then the same sequence in reverse order with $\frac{1}{q}$ as the ratio is also a solution.) We get $a_{1}=1, n_{1}=5$, and $a_{2}=16, n_{2}=5$.

Thus, the solutions to the problem are the following two 5-term sequences (which can easily be verified to satisfy the conditions): $1,-2,4,-8,16$, and $16,-8,4,-2,1$.