4.063. Given two geometric progressions consisting of the same number of terms. The first term and the common ratio of the first progression are 20 and $3 / 4$, respectively, while the first term and the common ratio of the second progression are 4 and $2 / 3$, respectively. If the terms of these progressions with the same indices are multiplied, the sum of all such products is 158.75. Find the number of terms in this progression.
## Solution.

Let $a_{1}, a_{2}, a_{3}, \ldots$ and $b_{1}, b_{2}, b_{3}, \ldots$ be the terms of two geometric progressions, where $a_{1}=20, q_{1}=\frac{3}{4}; b_{1}=4, q_{2}=\frac{2}{3}$ and $a_{1} b_{1}+a_{2} b_{2}+a_{3} b_{3}+\ldots=$ $=158.75$. We find $a_{2}=a_{1} q_{1}=20 \cdot \frac{3}{4}=15, a_{3}=a_{1} q_{1}^{2}=20 \cdot \frac{9}{16}=\frac{45}{4}, b_{2}=b_{1} q_{2}=$ $=4 \cdot \frac{2}{3}=\frac{8}{3}, b_{3}=b_{1} q_{2}^{2}=4 \cdot \frac{4}{9}=\frac{16}{9}$. Then $20 \cdot 4+15 \cdot \frac{8}{3}+\frac{45}{4} \cdot \frac{16}{9}+\ldots=158.75$ or $80+40+20+\ldots=\frac{635}{4}$. The left side of this equation is the sum of the terms of a geometric progression, where the first term is 80 and the common ratio is $\frac{1}{2}$.

From this, we have

$$
\begin{aligned}
& \frac{80\left(1-\left(\frac{1}{2}\right)^{n}\right)}{1-\frac{1}{2}}=\frac{635}{4}, \frac{160\left(2^{n}-1\right)}{2^{n}}=\frac{635}{4}, \frac{32\left(2^{n}-1\right)}{2^{n}}=\frac{127}{4} \\
& 128\left(2^{n}-1\right)=127 \cdot 2^{n}, 2^{n}=2^{7}, n=7
\end{aligned}
$$

Answer: 7.