Seven, insert numbers between 55 and 555 so that they form an arithmetic sequence. The last number inserted equals the coefficient of the $x^{3}$ term in the expansion of $\left(\sqrt{\bar{x}}+\frac{1}{x}\right)^{1}$.

Keep the original text's line breaks and format, and output the translation result directly.
Seven, Solution: The general term formula of the expansion of $\left(\sqrt{\mathbf{x}}+\frac{1}{\mathbf{x}}\right)^{15}$ is
$$
\begin{array}{l}
T_{r+1}=C_{15}^{x}(\sqrt{\mathbf{x}})^{15-x}\left(\frac{1}{x}\right)^{\prime} \\
=C_{15}^{2} x^{\frac{15-3 r}{2}}, \\
\text { Let } \frac{15-3 r}{2}=3, \quad \text { then } r=3 . \\
\therefore C_{15}^{3}=\frac{15 \times 14 \times 13}{1 \times 2 \times 3}=455 .
\end{array}
$$

In an arithmetic sequence, $\mathrm{a}_{\mathrm{a}}=555, \mathrm{a}_{\mathrm{a}-1}=455$,
Therefore, $\mathrm{d}=100$.
By $555=55+(n-1) \times 100$,
Solving for $\mathrm{n}$ gives $\mathrm{n}=6$.
$\therefore$ Inserting 4 numbers between 55 and 555 makes them form an arithmetic sequence.