## Task A-4.2.

Five mutually distinct real numbers $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ are consecutive terms of an arithmetic sequence, and their sum is 50. Determine these numbers if the numbers $a_{1}, a_{2}$, and $a_{5}$ are consecutive terms of a geometric sequence.
## Solution.

Since $a_{1}, a_{2}, a_{3}, a_{4}$ and $a_{5}$ are consecutive terms of an arithmetic sequence, there exists a real number $d$ such that $a_{2}=a_{1}+d, a_{3}=a_{1}+2 d, a_{4}=a_{1}+3 d$ and $a_{5}=a_{1}+4 d$.

Given $50=a_{1}+a_{2}+a_{3}+a_{4}+a_{5}=5 a_{1}+10 d=5\left(a_{1}+2 d\right)$, we get $a_{1}+2 d=10$.

Since $a_{1}$, $a_{2}$ and $a_{5}$ are consecutive terms of a geometric sequence, we have

$$
a_{1} \cdot a_{5}=a_{2}^{2}
$$

or

$$
a_{1} \cdot\left(a_{1}+4 d\right)=\left(a_{1}+d\right)^{2}
$$

If we substitute $a_{1}+2 d=10$ into this equation, we get $(10-2 d)(10+2 d)=(10-d)^{2} . \quad 1$ point

Solving the above equation, we get $d=0$ or 4.

1 point

Since $a_{1}, a_{2}, a_{3}, a_{4}$ and $a_{5}$ are distinct real numbers, the case $d=0$ is ruled out, so we conclude that $d=4$ is necessary.

1 point

Substituting, we get the required numbers $a_{1}=2, a_{2}=6, a_{3}=10, a_{4}=14 \text{ and } a_{5}=18$.

1 point