8.3.1. (12 points) Five numbers form an increasing arithmetic progression. The sum of their cubes is zero, and the sum of their squares is 70. Find the smallest of these numbers.
Answer: $-2 \sqrt{7} \approx-5.29$.

Solution. Let's represent the terms of the progression as

$$
a-2 d, a-d, a, a+d, a+2 d
$$

Then the sum of the squares is

$$
(a-2 d)^{2}+(a-d)^{2}+a^{2}+(a+d)^{2}+(a+2 d)^{2}=5 a^{2}+10 d^{2}=70
$$

and the sum of the cubes is

$$
(a-2 d)^{3}+(a-d)^{3}+a^{3}+(a+d)^{3}+(a+2 d)^{3}=5 a^{3}+30 a d^{2}=a\left(5 a^{2}+30 d^{2}\right)=0 .
$$

From this, $a=0, d=\sqrt{7}$, and the smallest term of the sequence $a-2 d=-2 \sqrt{7}$.