An arithmetic sequence is a sequence in which each term after the first is obtained by adding a constant to the previous term.

If the first four terms of an arithmetic sequence are $a, 2 a, b$, and $a-6-b$ for some numbers $a$ and $b$, then the value of the 100th term is
(A) -100
(B) -300
(C) 150
(D) -150
(E) 100
Since each term is obtained by adding the same number to the previous term, then the differences between pairs of consecutive terms are equal.

Looking at the first three terms, we thus have $2 a-a=b-2 a$ or $b=3 a$.

Therefore, in terms of $a$, the first four terms are $a, 2 a, 3 a$, and $a-6-3 a=-6-2 a$.

Since the constant difference between the terms equals $a$ (as $2 a-a=a$ ), then the fourth term should be $4 a$, so $4 a=-6-2 a$ or $6 a=-6$ or $a=-1$.

Thus, the sequence begins $-1,-2,-3,-4$.

The 100th term is thus -100 (which we can get by inspection or by saying that we must add the common difference 99 times to the first term, to get $-1+99(-1)=-100)$.

ANswer: (A)