3-ча 1. Construct two progressions: an arithmetic and a geometric one, each consisting of four terms; in such a way that if the corresponding terms of both progressions are added, the resulting numbers should be: $27, 27, 39, 87$.
Let $a, a+d, a+2 d, a+3 d$ be the desired arithmetic progression, and $b, b q, b q^{2}, b q^{3}$ be the desired geometric progression. According to the problem,

\[
\begin{aligned}
a+b & =27 \\
a+d+b q & =27 \\
a+2 d+b q^{2} & =39 \\
a+3 d+b q^{3} & =87
\end{aligned}
\]

Subtract the first equation from the second, the second from the third, and the third from the fourth:

\[
\begin{aligned}
d+b(q-1) & =0 \\
d+b q(q-1) & =12 \\
d+b q^{2}(q-1) & =48
\end{aligned}
\]

From the first equation, we get $b(q-1)=-d$; substitute this expression into the second and third equations:

\[
\begin{aligned}
d-d q & =12 \\
d-d q^{2} & =48
\end{aligned}
\]

Dividing the last equation by the second-to-last equation, we get $q=3$. Therefore, $d=-6, b=3$ and $a=24$. Thus, the desired progressions are

\[
\begin{array}{cccc}
24, & 18, & 12, & 6 \\
3, & 9, & 27, & 81
\end{array}
\]