3. If the expansion of $(a+2 b)^{n}$ has three consecutive terms whose binomial coefficients form an arithmetic sequence, then the largest three-digit positive integer $n$ is $\qquad$ .
3. 959 .

Let the binomial coefficients $\mathrm{C}_{n}^{k-1}, \mathrm{C}_{n}^{k}, \mathrm{C}_{n}^{k+1}(1 \leqslant k \leqslant n-1)$ of three consecutive terms in the expansion of $(a+2 b)^{n}$ satisfy
$$
\begin{array}{l}
2 \mathrm{C}_{n}^{k}=\mathrm{C}_{n}^{k-1}+\mathrm{C}_{n}^{k+1} . \\
\text { Then } n^{2}-(4 k+1) n+4 k^{2}-2=0 \\
\Rightarrow n_{1,2}=\frac{4 k+1 \pm \sqrt{8 k+9}}{2} .
\end{array}
$$

Since $n$ is a positive integer, let
$$
8 k+9=(2 m+1)^{2} \Rightarrow 2 k=m^{2}+m-2 \text {. }
$$

Substituting into equation (1) gives
$$
n_{1}=(m+1)^{2}-2, n_{2}=m^{2}-2 \text {. }
$$

Therefore, the maximum value of the three-digit positive integer $n$ is 959 .