6. Given a finite sequence that satisfies: the sum of any 3 consecutive terms is negative, and the sum of any 4 consecutive terms is positive, then the maximum number of terms in this sequence is
A. 7
B. 6
C. 5
D. 4
Construct an array as shown below, with each row consisting of positive numbers, then the sum of 12 items is positive;
$$
\begin{array}{llll}
a_{1} & a_{2} & a_{3} & a_{4}
\end{array}
$$

and each column consisting of negative numbers, then the sum of 12 items is negative, which is a contradiction.
$$
a_{2} \quad a_{3} \quad a_{4} \quad a_{5}
$$

Therefore, the sequence cannot exceed 5 terms. When a 5-term sequence satisfies the condition,
$$
a_{3} \quad a_{4} \quad a_{5} \quad a_{6}
$$

we have $a_{1}+a_{2}+a_{3}+a_{4}>0 \Rightarrow a_{1}>0, a_{4}>0$; similarly, we get $a_{2}>0, a_{5}>0$, thus constructing the sequence $2,2,-5,2,2$ that satisfies the condition. Therefore, the answer is $C$.