4.073 Find the sum of the first four terms of a geometric progression that has the property that its first three terms, the sum of which is $\frac{148}{9}$, are simultaneously the first, fourth, and eighth terms of some arithmetic progression.
Solution. Let $b, b q, b q^{2}, b q^{3}$ be the given geometric progression, then

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-087.jpg?height=179&width=1050&top_left_y=873&top_left_x=194)

(if $q=1$, then $b q$ and $b q^{2}$ would not necessarily be the fourth and eighth terms of the arithmetic progression) $\Rightarrow b+b q+b q^{2}+b q^{3}=b \cdot \frac{q^{4}-1}{q-1}=\frac{700}{27}=25 \frac{25}{27}$.

Answer: $25 \frac{25}{27}$.