6. A finite sequence satisfies: the sum of any three consecutive terms is negative, and the sum of any four consecutive terms is positive. Then the maximum number of terms in this sequence is ( ).
(A) 7
(B) 6
(C) 5
(D) 4
6. C.

On the one hand, a sequence of five terms can be constructed: $2, 2, -5, 2, 2$, which meets the conditions;

On the other hand, it can be proven that a sequence satisfying the conditions cannot have more than five terms.
Otherwise, take the first six terms and arrange them as follows:
$$
\begin{array}{l}
a_{1}, a_{2}, a_{3} \\
a_{2}, a_{3}, a_{4} \\
a_{3}, a_{4}, a_{5} \\
a_{4}, a_{5}, a_{6}
\end{array}
$$

Since the sum of each row is negative, the sum of these 12 numbers is negative; since the sum of each column is positive, the sum of these 12 numbers is positive. This is a contradiction.