A convex quadrilateral's two opposite sides are each divided into 100 equal parts, and the points of the same ordinal number are connected. Among the 100 quadrilaterals thus obtained, the area of the first one is 1, and the area of the last, the 100th one, is 2 square units. What is the area of the original quadrilateral?
We will show that the areas of the small quadrilaterals form an arithmetic sequence. Using this, we can easily calculate the area of the large quadrilateral.

Let the opposite sides of the original quadrilateral, which is divided, be $AB$ and $CD$, and cut the small quadrilaterals into triangles along one of their diagonals as shown in Figure 1 - so that the drawn diagonals do not have a common point.

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Figure 1

Our statement would follow from the fact that the areas of the triangles based on $AB$ and similarly on $CD$ form an arithmetic sequence, since in this way we get the area of the small quadrilaterals as the sum of the corresponding elements of an arithmetic sequence.

The sides of the triangles in question that lie on $AB$ are equal according to the condition, so it is sufficient to show that the heights corresponding to this side - the perpendiculars dropped from the division points of $CD$ to the opposite $AB$ side - form an arithmetic sequence.

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1987-05-216-2 . \mathrm{eps} \\
\text { Figure 2 }
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Consider three adjacent division points on the $CD$ side and the three corresponding heights (Figure 2). In the trapezoid where the two "outer" heights are the bases, the "middle" height is the midline, so its length is the arithmetic mean of the bases. This precisely means that the heights in question indeed form an arithmetic sequence.

The area $T$ of the original quadrilateral is thus the sum of the first 100 terms of an arithmetic sequence, the first and hundredth elements of which we know. The sum of the first 100 elements is therefore

$$
T=\frac{a_{1}+a_{100}}{2} \cdot 100=150 \text { units. }
$$

Remark: In the solution, we used two fundamental properties of arithmetic sequences, which can be easily proven from the definition.

On the one hand, if $u_{n}$ and $v_{n}$ are two arithmetic sequences, and $a$ and $b$ are any real numbers, then $a \cdot u_{n}+b \cdot v_{n}$ is also an arithmetic sequence. On the other hand, $u_{n}$ is an arithmetic sequence if and only if for any three consecutive elements, the middle one is the arithmetic mean of the two outer ones.