## Task A-4.6.

Let $a_{1}, a_{2}, \ldots, a_{41}$ be an arithmetic sequence such that

$$
\frac{1}{\sqrt{a_{1}}+\sqrt{a_{2}}}+\frac{1}{\sqrt{a_{2}}+\sqrt{a_{3}}}+\cdots+\frac{1}{\sqrt{a_{40}}+\sqrt{a_{41}}}
$$

is a natural number. If $a_{1}=1$ and the difference of the sequence is a natural number, determine the difference of the sequence.
Scoring: Rationalizing the fractions in the given sum carries 3 points, 1 point is for addition, and another 1 point is for writing the sum using the required variable $d$ (the difference of the sequence) and the fact that $a_{1}=1$. This part is common in both provided solutions.

Furthermore, in the first solution, the conclusion that $\sqrt{40 d+1}=k$ is a natural number carries 1 point, and the further 2 points are for concluding that $\frac{k^{2}-1}{40}$ and $\frac{40}{k+1}$ are natural numbers. 2 points are for calculating the possible values for $k$, and the final 1 point is for calculating the possible differences of the sequence.

In the second solution, we continue by squaring and conveniently rearranging the obtained expression, which carries 2 points, and determining the possible options for $n$ carries 2 points. The student can simply substitute different values for $n$ until they get $40-2 n<n^{2}$ and state that this will hold for all larger $n$, for which they should also receive 2 points. The final 1 point is for substituting the possible values for $n$ and finding all possible differences of the sequence.