The first term of an arithmetic sequence is $e$, and the last term is $u$. Can a given number $k$ that lies between $e$ and $u$ be a term of this sequence? If so, what could be the number of terms in the sequence? - Can the sequence be 100 terms long, where
a) $e=1, \quad u=1000, \quad k=343$
b) $\quad e=81 \sqrt{2}-64 \sqrt{3}, \quad u=54 \sqrt{2}-28 \sqrt{3}, \quad k=69 \sqrt{2}-48 \sqrt{3}$?
Let the difference of the sequence be $d$, the number of terms be $n$, and on the other hand, suppose that $k$ is a term of the sequence, and its position is $m$. Then clearly $1k<u$; and

$$
\begin{gathered}
u=e+(n-1) d, \quad k=e+(m-1) d, \quad \text { so } \\
d=\frac{u-e}{n-1}=\frac{k-e}{m-1}
\end{gathered}
$$

thus, gathering the given numbers on the right side,

$$
\frac{m-1}{n-1}=\frac{k-e}{u-e}
$$

Here, the left side must be the ratio of two natural numbers, so $k$ can only be a term of the sequence if the ratio $(k-e) /(n-e)$ is also (less than 1) a positive rational number. (Given the assumption, this ratio is certainly between 0 and 1, but not necessarily rational.)

Conversely, if $e \lessgtr k \lessgtr u$, and simplifying

$$
\frac{k-e}{u-e}=\frac{p}{q}
$$

where $p, q$ are relatively prime natural numbers, $p<q$, then there can be - and indeed infinitely many - arithmetic sequences for which $e, k$ and $u$ are terms. Indeed, the corresponding $m, n$ number pair can be given:

$$
\frac{m-1}{n-1}=\frac{k-e}{u-e}=\frac{p}{p}=\frac{t p}{t q}
$$

from which $m=t p+1, n=t q+1$, where $t$ is any natural number.

In the given numerical examples, the number of terms is also given, so from $t q+1=100$ we get $t q=99$, so the answer will only be affirmative if the value of the ratio $(k-e) /(u-e)$ is a rational number between 0 and 1, whose irreducible form has a denominator that is a divisor of 99 greater than 1: 3, 9, 11, 33, or 99. - This requirement is not met in the a) example: $(k-e) /(u-e)=342 / 999-38 / 111$, but it is met in the b) example, because

$$
\frac{k-e}{u-e}=\frac{16 \sqrt{3}-12 \sqrt{2}}{36 \sqrt{3}-27 \sqrt{2}}=\frac{4(4 \sqrt{3}-3 \sqrt{2})}{9(4 \sqrt{3}-3 \sqrt{2})}=\frac{4}{9}
$$

and $q=9$ is among the allowed denominators. (Since this ratio is between 0 and 1, it is not necessary to check whether $k$ lies between $e$ and $u$.) Indeed, from $u-e=99 d=36 \sqrt{3}-27 \sqrt{2}$ we get $d=(4 \sqrt{3}-3 \sqrt{2}) / 11$, and on the other hand

$$
\frac{4}{9}=\frac{44}{99}=\frac{m-1}{n-1}
$$

from which $m=45$, and thus the 45th term of the sequence is:

$$
e+44 d=(81 \sqrt{2}-64 \sqrt{3})+\frac{44}{11}(4 \sqrt{3}-3 \sqrt{2})=69 \sqrt{2}-48 \sqrt{3}=k
$$

Jenő Kisvölcsey (Budapest, Piarist g. IV. o. t.).