Constant sum

Let the given sequence be: $1, a, b, c, 5$ We now have the following system of equations: $\begin{aligned} 1 + a + b &= 7 \\ a + b + c &= 7 \\ b + c + 5 &= 7 \end{aligned}$

From the first equation, we see that $a+b=6$ . Substituting this into the second equation yields $c=1$ . Substituting that value into the third equation yields $b=1$ , which in turn implies that $a=5$ . So the sequence is $1,5,1,1,5$ and the number $\boxed{1}$ appears most often.

Since the second and third terms must add up to six, the fourth term must be a one.

This implies that the third term must also be a one (so that the last three add up to 5).

Therefore, since we have three ones already, $\boxed1$ appears the most.