8. For a real number $t$, it is known that the first three terms of the geometric sequence $\left\{a_{n}\right\}$ are $2 t, 5 t-1, 6 t+2$, and the sum of the first $n$ terms of this sequence is $S_{n}$. Then the maximum integer value of $n$ that satisfies the inequality $\left|S_{n}-\frac{1}{65}\right|<1$ is ( ).
(A) 2
(B) 3
(C) 5
(D) 8
8. A.

From the given, we have $(5 t-1)^{2}=2 t(6 t+2)$.
Solving, we get $t=1$ or $t=\frac{1}{13}$. When $t=1$, $S_{n}=\frac{2\left(1-2^{n}\right)}{1-2}=2^{n+1}-2$. Then, for $\left|S_{n}-\frac{1}{65}\right|=\left|2^{n+1}-2-\frac{1}{65}\right|<1$, there is no positive integer $n$ that satisfies this condition.
When $t=\frac{1}{13}$,
$$
S_{n}=\frac{\frac{2}{13}\left[1-(-4)^{n}\right]}{1-(-4)}=\frac{2}{65}\left[1-(-4)^{n}\right] \text {. }
$$

From $\left|S_{n}-\frac{1}{65}\right|<1$, we get
$$
\left|1-2 \times(-4)^{n}\right|<65 \text {. }
$$

Solving, we get $-32<(-4)^{n}<33$.
Thus, the maximum value of $n$ is 2.