4.028. Find the number of terms in a finite geometric progression, where the first, second, and last terms are 3, 12, and 3072, respectively.
Solution.

From the condition we have $b_{1}=3, b_{2}=12, \ldots, b_{n}=3072$.

By formula (4.6) we get

$$
\left\{\begin{array} { l } 
{ b _ { 1 } = 3 , } \\
{ b _ { 1 } q = 1 2 , } \\
{ b _ { 1 } q ^ { n - 1 } = 3 0 7 2 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
b_{1}=3, \\
q=4, \\
4^{n-1}=1024
\end{array} \Rightarrow 4^{n-1}=4^{5} \Leftrightarrow n-1=5 \Rightarrow n=6\right.\right.
$$

Answer: 6.