2. Four numbers $a_{1}, a_{2}, a_{3}, a_{4}$ form an arithmetic progression. The product of the first and fourth terms is equal to the larger root of the equation

$$
x^{1+\log x}=0.001^{-\frac{2}{3}}
$$

and the sum of the squares of the second and third terms is five times the number of the term in the expansion of

$$
\left(\frac{1}{\sqrt[4]{y}}+\sqrt[6]{y}\right)^{20}
$$

which does not depend on $y$. Find these numbers.
Solution. To solve the given equation, we can rewrite it in the form

$$
x^{1+\log x}=10^{2}
$$

From which we obtain

$$
(1+\log x) \log x=2
$$

i.e., $\log x=1$ and $\log x=-2$ or $x=10$ and $x=10^{-2}$.

Now, let's find the term in the expansion of the given binomial that does not depend on $y$. The $(k+1)$-th term is:

$$
\binom{20}{k} y^{\frac{k-20}{4}+\frac{k}{6}}
$$

and if it does not depend on $y$, it must be

$$
\frac{k-20}{4}+\frac{k}{6}=0
$$

i.e., $k=12$. Thus, the 13th term does not depend on $y$.

From the condition of the problem, we have

$$
a_{1} a_{4}=10, \quad a_{2}^{2}+a_{3}^{2}=5 \cdot 13
$$

i.e.,

$$
a_{1}\left(a_{1}+3 d\right)=10, \quad 2 a_{1}^{2}+5 d^{2}+6 a_{1} d=65
$$

from which we get $d= \pm 3, a_{1}=\mp 10$. Thus, the required numbers are $-10,-7,-4,-1$ or $10,7,4,1$.