4.066. In a certain geometric progression containing $2 n$ positive terms, the product of the first term and the last is 1000. Find the sum of the decimal logarithms of all the terms of the progression.
## Solution.

Let $b_{1}, b_{2}, b_{3}, \ldots, b_{n}$ be the terms of a geometric progression, the number of which is $2n$, and $b_{1} \cdot b_{n}=1000$. We have

$$
\begin{aligned}
& X=\lg b_{1}+\lg b_{2}+\lg b_{3}+\ldots+\lg b_{n}=\lg \left(b_{1} \cdot b_{2} \cdot b_{3} \ldots b_{n}\right)= \\
& =\lg \left(\left(b_{1} \cdot b_{n}\right) \cdot\left(b_{2} \cdot b_{n-1}\right) \cdot\left(b_{3} \cdot b_{n-2}\right) \ldots\right)
\end{aligned}
$$

Since $b_{k} \cdot b_{m}=b_{p} b_{q}$, where $k+m=p+q$, then

$$
X=\lg \left(b_{1} b_{n}\right)^{n}=\lg 1000^{n}=n \lg 1000=3 n
$$

Answer: $3 n$.