Problem 861

Given the task,

$$a_{1}\left(1+q^{2}+q^{4}\right)=63 \quad \text { and } \quad a_{1}\left(q+q^{3}\right)=30$$

eliminating $a_{1}$ from these two equations, we get

$$30 q^{4}-63 q^{3}+30 q^{2}-63 q+30=0$$

dividing every term of the equation by $3 q^{2}$:

$$10\left(q^{2}+\frac{1}{q^{2}}\right)-21\left(q+\frac{1}{q}\right)+10=0$$

Let

$$q+\frac{1}{q}=y$$

then

$$q^{2}+\frac{1}{q^{2}}=y^{2}-2$$

so from (1) we get

$$10 y^{2}-21 y-10=0$$

from which

$$y_{1}=\frac{5}{2} \text { and } y_{2}=-\frac{2}{5}$$

Substituting these values into (2):

$$q_{1}=2, q_{2}=\frac{1}{2}, q_{3}=\frac{-1+2 i \sqrt{6}}{5}, q_{4}=-\frac{1+2 i \sqrt{6}}{5}$$

Thus, the geometric progressions consisting of real terms are

$3,6,12,24,48$ and $48,24,12,6,3$.

(Irén Juvancz, Nyíregyháza.)

The problem was also solved by: Boros J., Csete A., Dolowschiák M., Filkorn J., Goldberger M., Kárf J., Kornis Ö., Krausz B., Krisztián Gy., Lukhaub Gy., Mandel M., Miliczer L., Obláth R., Petrogalli G., Pollák N., Porkoláb J., Prakatur T., Romsauer Etta., Róth D., Sasvári G., Sasvári J., Schieb Á., Spitzer Ö., Weisz J.