An algebra problem by A Former Brilliant Member

Relevant wiki: Arithmetic Progressions

Let $a$ be the first term and $d$ be the common difference. Then the four positive integers are $a,a+d,a+2d$ and $a+3d$ . The sum of these four positive integers is $4a+6d=46$ , so, $2a+3d=23$ . Solving for $d$ , we find $d=\dfrac{23-2a}{3}$ .

The third term is $a+2d=a+2 \left(\dfrac{23-2a}{3}\right)=\dfrac{46-a}{3}$

Thus, to maximize this expression, we should minimize $a$ . Since $a$ is a positive integer, the smallest possible value of $a$ is $1$ . It follows that

$d=\dfrac{23-2}{3}=7$

which gives us an arithmetic sequence of $1,8,15,22$ .

Therefore, the greatest possible third term is $15$ .