The sum of the first two terms of a geometric sequence is 80, and the sum of the first six terms is 665. What is the sum of the first four terms of the sequence?
Let the $n$-th term of the geometric sequence be denoted by $a_{n}$, and its common ratio by $q$. Consider the sequence $\left(b_{n}\right)$ whose terms are:

$$
b_{n}=a_{2 n-1}+a_{2 n} \quad(n=1,2,3)
$$

The sequence $\left(b_{n}\right)$ is also a geometric sequence with a common ratio of $q^{2}$. According to the problem,

$$
b_{1}=a_{1}+a_{2}=80
$$

$$
b_{1}+b_{2}+b_{3}=b_{1}\left(1+q^{2}+q^{4}\right)=665
$$

Therefore,

$$
1+q^{2}+q^{4}=\frac{665}{80}
$$

which simplifies to

$$
16 q^{4}+16 q^{2}-117=0
$$

From this, $q^{2}=\frac{9}{4}$ is obtained, since $q^{2}$ cannot be negative.

The desired sum is:

$$
a_{1}+a_{2}+a_{3}+a_{4}=b_{1}+b_{2}=b_{1}\left(1+q^{2}\right)=80\left(1+\frac{9}{4}\right)=260
$$

Two sequences satisfy the conditions of the problem. One is: $32 ; 48 ; 72 ; 108 ; 162 ; 243$. The other is: $-160 ; 240 ;-360 ; 540$; $-810 ; 1215$.

Bagyinszky Róbert (Békéscsaba, Széchenyi I. Közg. Szki., IV. o. t.)