An arithmetic progression's first, second, fifth, and last terms are consecutive terms of a geometric progression, the sum of which is 80. Let's find these two progressions.
If the first term of an arithmetic progression is $a$, the difference is $d$, and the number of terms is $n$, then the numbers

$$
a, a+d, a+4 d \text{ and } a+(n-1) d
$$

form a geometric progression according to the problem, hence

$$
(a+d)^{2}=a(a+4 d)
$$

from which

$$
d=2 a
$$

Substitute the value of $d$ just found into the geometric progression:

$$
a, 3 a, 9 a, a+(n-1) 2 a
$$

from which it follows that

$$
1+2(n-1)=27
$$

or

$$
n=14
$$

But

$$
4 a+(n+4) d=4 a+18 d=40 a=80
$$

so

$$
a=2, d=4
$$

The two progressions themselves are:

$2,6,10, \ldots 54$

and

$2,6,18, \ldots 54$

(Preisich Gusztáv, Beszterczebánya.)

The problem was also solved by: Ádámffy E., Bartók I., Deutsch E., Deutsch I., Eckstein J., Enyedi B, Glück J., Haar A., Harsányi Z., Hirschfeld Gy., Jánosy Gy., Kertész G., Kiss J., Korény Gy., Kürti I., Liebner A., Neidenbach E., Pám M., Pichler S., Popoviciu M., Rássy P., Riesz K., Riesz M., Sárközy E., Schlesinger O., Schuster Gy., Schwemmer I., Szántó H., Szőke D., Szücs A., Tóth B., Weisz P.