4.045. The sum of the first three terms of a geometric progression is 91. If 25, 27, and 1 are added to these terms respectively, the resulting three numbers form an arithmetic progression. Find the seventh term of the geometric progression.
## Solution.

From the condition, we have:

$b_{1}, b_{1} q, b_{1} q^{2}$ - terms of a geometric progression,

$b_{1}+25, b_{1} q+27, b_{1} q^{2}$ - terms of an arithmetic progression, then we get

$$
\left\{\begin{array} { l } 
{ b _ { 1 } + b _ { 1 } q + b _ { 1 } q ^ { 2 } = 9 1 , } \\
{ 2 ( b _ { 1 } q + 2 7 ) = b _ { 1 } + 2 5 + b _ { 1 } q ^ { 2 } + 1 , }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
b_{1}\left(1+q+q^{2}\right)=91 \\
b_{1}\left(q^{2}-2 q+1\right)=28
\end{array} \Rightarrow\right.\right.
$$

$$
\Rightarrow b_{1}=\frac{91}{1+q+q^{2}}, \frac{91\left(q^{2}-2 q+1\right)}{1+q+q^{2}}=28,3 q^{2}-10 q+3=0, \text { from which }
$$

$q_{1}=3, q_{2}=\frac{1}{3}$

Then $b_{1}^{\prime}=\frac{91}{1+3+9}=\frac{91}{13}=7$ or

$$
b_{1}^{\prime \prime}=\frac{91}{1+\frac{1}{3}+\frac{1}{9}}=\frac{819}{9+3+1}=\frac{819}{13}=63
$$

From this, $b_{7}^{\prime}=b_{1}^{\prime} \cdot q_{1}^{6}=7 \cdot 3^{6}=5103 ; b_{7}^{\prime \prime}=b_{1}^{\prime \prime} q_{2}^{6}=63 \cdot \frac{1}{3^{6}}=\frac{63}{729}=\frac{7}{81}$.

Answer: 5103 or $\frac{7}{81}$.