1. The first term of an infinite geometric sequence with a sum of 3 is a natural number, and the quotient is equal to the reciprocal of an integer. Find the first term and the quotient of this sequence.
IV/1. Let the first term $a_{1}=n$ and the common ratio $q=\frac{1}{m}$. Since the sum of the geometric series is finite, $|q|<1$. Then $n+n q+n q^{2}+\cdots=\frac{n}{1-q}=\frac{n}{1-\frac{1}{m}}=3$. From this, we get $n=3\left(1-\frac{1}{m}\right)=3-\frac{3}{m}$. Since $n$ is a natural number, there are only two possibilities for $m$, namely $m_{1}=3$ and $m_{2}=-3$. Thus, we have two solutions to the problem: the sequence with the first term $a_{1}=2$ and common ratio $q=\frac{1}{3}$, or the sequence with the first term $a_{1}=4$ and common ratio $q=-\frac{1}{3}$.

![](https://cdn.mathpix.com/cropped/2024_06_07_a8720c28cfa2c07912d2g-13.jpg?height=68&width=1633&top_left_y=497&top_left_x=220)

![](https://cdn.mathpix.com/cropped/2024_06_07_a8720c28cfa2c07912d2g-13.jpg?height=54&width=1631&top_left_y=550&top_left_x=218)

The expression $n=3-\frac{3}{m}$ (or similar) ...........................................................................

![](https://cdn.mathpix.com/cropped/2024_06_07_a8720c28cfa2c07912d2g-13.jpg?height=54&width=1633&top_left_y=647&top_left_x=220)

Solutions $a_{1}=2, q=\frac{1}{3}$ and $a_{1}=4, q=-\frac{1}{3} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$ 1 point each