3. (1994 National High School Competition Question) Given the sequence $\left\{a_{n}\right\}$ satisfies $3 a_{n+1}+a_{n}=4(n \geqslant 1)$ and $a_{1}=9$, the sum of the first $n$ terms is $S_{n}$, then the smallest integer $n$ that satisfies the inequality $\left|S_{n}-n-6\right|<\frac{1}{125}$ is ( ).
A. 5
B. 6
C. 7
D. 8
3. C. The original recurrence relation is transformed into $3\left(a_{n+1}-1\right)=-\left(a_{n}-1\right)$ (where $x=1$ is the root of the equation $3 x+x=4$). Let $b_{n}=a_{n}-1$, then $b_{n+1}=-\frac{1}{3} b_{n}, b_{1}=a_{1}-1=8$, so $b_{n}$ is a geometric sequence with the first term 8 and common ratio $-\frac{1}{3}$. Therefore, $S_{n}-n=\left(a_{1}-1\right)+\left(a_{2}-2\right)+\cdots+\left(a_{n}-n\right)=b_{1}+b_{2}+\cdots+b_{n}=\frac{8\left[1-\left(-\frac{1}{3}\right)^{n}\right]}{1-\left(-\frac{1}{3}\right)}=6-6\left(-\frac{1}{3}\right)^{n}$. Thus, $\left|S_{n}-n-6\right|=6\left(\frac{1}{3}\right)^{n}<250$. The smallest positive integer $n$ that satisfies this inequality is $n=7$, hence the answer is C.