4.062. Represent the number 180 as the sum of four terms so that they form a geometric progression, in which the third term is 36 more than the first.
Solution.

From the condition

$$
\begin{aligned}
& \left\{\begin{array} { l } 
{ b _ { 1 } + b _ { 2 } + b _ { 3 } + b _ { 4 } = 1 8 0 , } \\
{ b _ { 3 } = b _ { 1 } + 3 6 , }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
b_{1}+b_{1} q+b_{1} q^{2}+b_{1} q^{3}=180, \\
b_{1} q^{2}=b_{1}+36,
\end{array} \Leftrightarrow\right.\right. \\
& \Leftrightarrow\left\{\begin{array}{l}
b_{1}\left(1+q+q^{2}+q^{3}\right)=180, \\
b_{1}\left(q^{2}-1\right)=36 .
\end{array} \Rightarrow \frac{b_{1}\left(q^{3}+q^{2}+q+1\right)}{b_{1}\left(q^{2}-1\right)}=\frac{180}{36}\right. \\
& \frac{q^{2}(q+1)+(q+1)}{(q+1)(q-1)}=5, \frac{(q+1)\left(q^{2}+1\right)}{(q+1)(q-1)}=5, \frac{q^{2}+1}{q-1}=5, q^{2}-5 q+6=0
\end{aligned}
$$

Substituting the value $q_{1}=3$ into the second equation of this system, we get $b_{1}^{\prime}=\frac{36}{q^{2}-1}=\frac{36}{8}=\frac{9}{2}$.

Then

$$
b_{2}^{\prime}=b_{1} q=\frac{9}{2} \cdot 3=\frac{27}{2}, b_{3}^{\prime}=b_{1} q^{2}=\frac{9}{2} \cdot 9=\frac{81}{2}, b_{4}^{\prime}=b_{1} q^{3}=\frac{9}{2} \cdot 27=\frac{243}{2} .
$$

Substituting the value $q_{2}=2$ into the second equation of this system, we get

$$
\begin{aligned}
& b_{1}^{n}=\frac{36}{q^{2}-1}=\frac{36}{4-1}=12, \quad b_{2}^{\prime \prime}=b_{1} q=12 \cdot 2=24, \quad b_{3}^{\prime \prime}=b_{1} q^{2}=12 \cdot 4=48 \\
& b_{4}^{\prime \prime}=b_{1} q^{3}=12 \cdot 8=96
\end{aligned}
$$

Answer: 1) $\left.\frac{9}{2}+\frac{27}{2}+\frac{81}{2}+\frac{243}{2} ; 2\right) 12+24+48+96$.