The nth term of a certain sequence, for all $n>2$, can be obtained by subtracting from the previous term the term before that. If the first term is 4 and the second term is 8, what is the sum of the first 100 terms of the sequence?

This one tripped me up at first because I misread 'the term before that' as the term before the previous term. Once I got the pattern, it was smooth.

I listed out the first few terms: 4, 8, 4, -4, -8, -4, 4, 8... and noticed it repeats every 6 terms. Then I just found the sum of one cycle and multiplied by 100/6. But 100 isn't divisible by 6, so I had to handle the remainder. A bit tricky but doable.

Is the answer 4? I got that after finding the cycle sum is 0 and then the remaining 4 terms sum to 4. But I'm not 100% sure.

This is a classic sequence problem. The recurrence is a_n = a_{n-1} - a_{n-2}. For n=3: a_3 = 8-4=4, n=4: a_4=4-8=-4, etc. The sequence is periodic with period 6: 4,8,4,-4,-8,-4, then repeats. Sum of one period is 0. 100 = 6*16 + 4, so sum of first 100 terms is sum of first 4 terms = 4+8+4-4=12. Wait, that gives 12, not 4. Let me recalculate: first 4 terms: 4,8,4,-4 sum = 12. So answer should be 12? But 12 is choice D. Hmm, but I thought cycle sum was 0, and 96 terms sum to 0, plus first 4 sum to 12. So 12. But earlier I said 4, that was wrong.