3. Let $n<100$. Then the largest integer $n$ such that the expansion of $(a+b)^{n}$ has three consecutive terms with coefficients in arithmetic progression is $\qquad$ . . .
3. 98.

Let the coefficients of three consecutive terms in the expansion of $(a+b)^{n}$ be $\mathrm{C}_{n}^{k-1}, \mathrm{C}_{n}^{k}, \mathrm{C}_{n}^{k+1} (1 \leqslant k \leqslant n-1)$.
By the problem, we have $2 \mathrm{C}_{n}^{k}=\mathrm{C}_{n}^{k-1}+\mathrm{C}_{n}^{k+1}$.

Expanding and rearranging according to the definition of combination numbers, we get
$$
n^{2}-(4 k+1) n+4 k^{2}-2=0 \text{. }
$$

Thus, $n_{1,2}=\frac{4 k+1 \pm \sqrt{8 k^{2}+9}}{2}\left(n_{1,2} \in \mathbf{N}_{+}\right)$,
$$
\begin{array}{l}
8 k^{2}+9=(2 m+1)^{2} \\
\Rightarrow 2 k=m^{2}+m-2 .
\end{array}
$$

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

From $(m+1)^{2}-2<100$, we know $n=98$.