A mean meaningful problem

Remember that if $a,b,c$ are in a 3-term arithmetic progression (hereafter abbreviated AP), then $a+c = 2b$ .

Observe that the mode will always be $2$ and the mean will always be $\frac{25+x}{7}$ . On the other hand, the median depends on the value of $x$ :

• Case 1: $x \le 2$

The median is $2$ . If two terms of an AP are equal, then all of them are; in this case, since the mode and the median are both $2$ , we have the mean $\frac{25+x}{7} = 2$ or $x = -11$ .

• Case 2: $2 \le x \le 4$

The median is $x$ , thus the terms are $2, x, \frac{25+x}{7}$ . There are three cases of the AP, considering the middle element:

• Case 2.1: $2$ is the middle element.

Thus $x + \frac{25+x}{7} = 2 \cdot 2$ , or $x = \frac{3}{8}$ . But this is not in the interval $[2,4]$ , so we discard it.

• Case 2.2: $x$ is the middle element.

Thus $2 + \frac{25+x}{7} = 2 \cdot x$ , or $x = 3$ .

• Case 2.3: $\frac{25+x}{7}$ is the middle element.

Thus $2 + x = 2 \cdot \frac{25+x}{7}$ , or $x = \frac{36}{5}$ . But this is not in the interval $[2,4]$ , so we discard it.

• Case 3: $x \ge 4$

The median is $4$ , thus the terms are $2, 4, \frac{25+x}{7}$ . There are three APs involving the terms $2, 4$ , namely $(0,2,4), (2,3,4), (2,4,6)$ , and thus three cases for the value of $\frac{25+x}{7}$ , namely $0, 3, 6$ . This gives $x = -25, -4, 17$ respectively. But only $x = 17$ satisfies $x \ge 4$ , so that's the only solution from this case.

In total, we obtained the solutions $-11, 3, 17$ . The maximum of these is $\boxed{17}$ .