4. Let the increasing sequence $\left\{a_{n}\right\}$ satisfy $a_{1}=6, n \in \mathbf{N}_{+}$, and when $n \geqslant 2$, $a_{n}+a_{n-1}=\frac{9}{a_{n}-a_{n-1}}+8$. Then $a_{x}=$ $\qquad$
4.29.
$$
\begin{array}{l}
\text { Given } a_{k}+a_{k-1}=\frac{9}{a_{k}-a_{k-1}}+8 \\
\Rightarrow a_{k}^{2}-a_{k-1}^{2}=8\left(a_{k}-a_{k-1}\right)+9 .
\end{array}
$$

Taking $k=2,3, \cdots, n$, summing up we get
$$
a_{n}^{2}-a_{1}^{2}=8\left(a_{n}-a_{1}\right)+9(n-1) \text {. }
$$

Thus, $a_{10}^{2}-6^{2}=8\left(a_{10}-6\right)+9 \times 69$, which means
$$
a_{70}^{2}-8 a_{70}-609=0 \text {. }
$$

Therefore, $a_{70}=29$ (negative value is discarded).