Let $p_{i}$ be the $i^{\text {th }}$ prime number; for example, $p_{1}=2, p_{2}=3$, and $p_{3}=5$. For each prime number, construct the point $Q_{i}\left(p_{i}, 0\right)$. Suppose $A$ has coordinates $(0,2)$. Determine the sum of the areas of the triangles $\triangle A Q_{1} Q_{2}, \triangle A Q_{2} Q_{3}, \triangle A Q_{3} Q_{4}, \triangle A Q_{4} Q_{5}, \triangle A Q_{5} Q_{6}$, and $\triangle A Q_{6} Q_{7}$.
Consider the diagram below:

![](https://cdn.mathpix.com/cropped/2024_04_17_a164435ebfb99ea5f1c4g-10.jpg?height=371&width=957&top_left_y=1774&top_left_x=644)

The sum of the areas of the triangles is equal to the area of the larger $\triangle A Q_{1} Q_{7}$.

This is because the triangles share vertex $A$, the base of $\triangle A Q_{2} Q_{3}$ begins where the base of $\triangle A Q_{1} Q_{2}$ ends, the base of $\triangle A Q_{3} Q_{4}$ begins where the base of $\triangle A Q_{2} Q_{3}$ ends, and so on, with all of $Q_{1}, Q_{2}, Q_{3}, Q_{4}, Q_{5}, Q_{6}$, and $Q_{7}$ lying on the $x$-axis.

All of $Q_{1}$ through $Q_{7}$ lie on the $x$-axis, so the height of $\triangle A Q_{1} Q_{7}$ is the $y$-coordinate of $A$, or 2 . The first seven prime numbers are $2,3,5,7,11,13$, and 17 , so $p_{1}=2$ and $p_{7}=17$. The base of $\triangle A Q_{1} Q_{7}$ has length $p_{7}-p_{1}=17-2=15$. Therefore, the sum of the areas of the triangles is $\frac{1}{2} \times 15 \times 2=15$.

ANSWER: 15