## Task 1 - 241231

Determine the first six terms $a_{1}, a_{2}, \ldots, a_{6}$ of all those sequences $\left(a_{n}\right)$ of real numbers that have the following properties (1) to (5):

(1) It holds that $a_{1}=-\frac{5}{2}$

(2) It holds that $a_{5}=3$.

(3) $a_{1}, a_{2}, a_{3}, a_{4}$ are (in this order) terms of an arithmetic sequence.

(4) $a_{4}, a_{5}, a_{6}$ are (in this order) terms of a geometric sequence.

(5) The sum of the first six terms of the sequence $\left(a_{n}\right)$ is $\frac{13}{2}$.
After (3), there is a real number $d$ such that $a_{2}=a_{1}+d, a_{3}=a_{1}+2 d$ and $a_{4}=a_{1}+3 d=3 d-\frac{5}{2}$. In particular,

$$
a_{1}+a_{2}+a_{3}+a_{4}=4 \cdot\left(-\frac{5}{2}\right)+6 d=6 d-10
$$

After (4), there is a real number $q$ such that $a_{5}=a_{4} \cdot q$ and $a_{6}=a_{5} \cdot q$. Because of (2), $a_{5}=3$, so $a_{6}=3 q$ and $a_{4}=3 d-\frac{5}{2}=\frac{3}{q}$. (Here, because $0 \neq 3=a_{5}=q \cdot a_{4}$, also $q \neq 0$.)

This implies $d=\frac{1}{3} \cdot\left(\frac{3}{q}+\frac{5}{2}\right)=\frac{1}{q}+\frac{5}{6}$.

Furthermore, $a_{5}+a_{6}=3+3 q$, so according to (6),

$$
\frac{13}{2}=a_{1}+\cdots+a_{6}=6 d-10+3+3 q=\frac{6}{q}+5-10+3+3 q=\frac{6}{q}-2+3 q
$$

or $3 q^{2}-\left(2+\frac{13}{2}\right) q+6=0$, thus $q^{2}-\frac{17}{6} q+2=0$, which leads to $q_{1 / 2}=\frac{17}{12} \pm \sqrt{\frac{289}{144}-\frac{288}{144}}=\frac{17 \pm 1}{12}$, hence $q_{1}=\frac{16}{12}=\frac{4}{3}$ and $q_{2}=\frac{18}{12}=\frac{3}{2}$.

In the case $q=q_{1}=\frac{4}{3}$, $a_{6}=3 q=4, a_{5}=3, a_{4}=\frac{3}{q}=\frac{9}{4}$ and $d=\frac{1}{q}+\frac{5}{6}=\frac{3}{4}+\frac{5}{6}=\frac{19}{12}$, so $a_{1}=-\frac{5}{2}$, $a_{2}=d-\frac{5}{2}=\frac{19-30}{12}=-\frac{11}{12}$ and $a_{3}=a_{2}+d=\frac{19-11}{12}=\frac{8}{12}=\frac{2}{3}$.

(Indeed, then

$a_{3}+d=\frac{8+19}{12}=\frac{27}{12}=\frac{9}{4}=a_{4} \quad$ and $\left.\quad a_{1}+\cdots+a_{6}=\frac{-30-11+8+27+36+48}{12}=\frac{78}{12}=\frac{13}{2}.\right)$

And in the case $q=q_{2}=\frac{3}{2}$, $a_{6}=3 q=\frac{9}{2}, a_{5}=3, a_{4}=2$ and $d=\frac{1}{q}+\frac{5}{6}=\frac{3}{2}$, so $a_{1}=-\frac{5}{2}, a_{2}=a_{1}+d=-1$ and $a_{3}=a_{2}+d=\frac{1}{2}$. Indeed, then $a_{3}+d=2=a_{4}$ and

$$
a_{1}+\cdots+a_{6}=\frac{-5-2+1+4+6+9}{2}=\frac{13}{2}
$$