The product of the first and the third terms of an arithmetic sequence is $5$. If all terms of the sequence are positive integers, what is the fourth term?
I know that an arithmetic sequence is a sequence of numbers where each term is obtained by adding a constant difference to the previous term. Let me call the first term $a$ and the difference $d$. Then the third term is $a + 2d$ and the fourth term is $a + 3d$. The problem says that $a(a + 2d) = 5$, which means that $a$ and $a + 2d$ are both positive factors of $5$. Since $5$ is a prime number, it has only two positive factors: $1$ and $5$. This means that either $a = 1$ and $a + 2d = 5$, or $a = 5$ and $a + 2d = 1$. But the second case is impossible, since it would imply that $d$ is negative, which contradicts the fact that all terms of the sequence are positive integers. So I can conclude that $a = 1$ and $a + 2d = 5$. Solving for $d$, I get $d = 2$. Therefore, the fourth term is $a + 3d = 1 + 3 \times 2 = 7$. 