10. Let $a_{n}>0, a_{1}=2$, and for $n \geqslant 2$, we have
$$
a_{n}+a_{n-1}=\frac{n}{a_{n}-a_{n-1}}+2 \text {. }
$$

Then the general term formula of the sequence $\left\{a_{n}\right\}$ is $a_{n}=$
10. $\sqrt{\frac{n(n+1)}{2}}+1$.

Since $a_{n}+a_{n-1}=\frac{n}{a_{n}-a_{n-1}}+2$, we have
$$
a_{n}^{2}-a_{n-1}^{2}=2\left(a_{n}-a_{n-1}\right)+n .
$$

Therefore, $\left(a_{n}-1\right)^{2}-\left(a_{n-1}-1\right)^{2}=n$.
Hence, $\left(a_{n}-1\right)^{2}$
$$
\begin{array}{l}
=\sum_{k=2}^{n}\left[\left(a_{k}-1\right)^{2}-\left(a_{k-1}-1\right)^{2}\right]+\left(a_{1}-1\right)^{2} \\
=\sum_{k=2}^{n} k+\left(a_{1}-1\right)^{2}=\sum_{k=1}^{n} k=\frac{n(n+1)}{2} .
\end{array}
$$

Thus, $a_{n}-1=\sqrt{\frac{n(n+1)}{2}}, a_{n}=\sqrt{\frac{n(n+1)}{2}}+1$.