$10 \cdot 64$ in the sequence $1,3,2, \cdots$, after the first two terms, each term is equal to the difference of the two preceding terms (the preceding term minus the term before it). The sum of the first 100 terms of this sequence is
(A) 5 .
(B) 4 .
(C) 2 .
(D) 1 .
(E) -1 .
(26th American High School Mathematics Examination, 1975)
[Solution] According to the problem, the first eight terms of this sequence are $1, 3, 2, -1, -3, -2, 1$, 3. Since the seventh and eighth terms are the same as the first and second terms, the ninth term will be the same as the third term, and so on; that is, this sequence repeats every 6 terms, i.e.,
$a_{k}=a_{k+6}(k=1,2,3, \cdots) . a_{k}$ is the general term of the sequence.
Also, $a_{1}+a_{2}+a_{3}+a_{4}+a_{5}+a_{6}=0$, so $\sum_{k=1}^{96} a_{k}=0$.
Therefore, $\sum_{k=1}^{100} a_{k}=\sum_{k=1}^{96} a_{k}+a_{97}+a_{98}+a_{99}+a_{100}$ $=0+1+3+2-1=5$.
Hence, the answer is $(A)$.