The GMAT loves sequence questions. Test-takers, not surprisingly, do not feel the same level of affection for this topic. In some ways, it’s a peculiar reaction. A sequence is really just a set of numbers. It may be infinite, it may be finite, but it’s this very open-endedness, this dizzying level of fuzzy abstraction, that can make sequences so difficult to mentally corral.

If you are one of the many people who fear and dislike sequences, your main consolation should come from the fact that the main weapon in the question writer’s arsenal is the very fear these questions might elicit. And if you have been a reader of this blog for any length of time, you know that the best way to combat this anxiety is to dive in and convert abstractions into something concrete, either by listing out some portion of the sequence, or by using the answer choices and working backwards.

Take this question for example:

*For a certain set of numbers, if x is in the set, then x – 3 is also in the set. If the number 1 is in the set, which of the following must also be in the set?** *

*I. **4*

*II. **-1*

*III. **-5*

*A) **I only*

*B) **II only*

*C) **III only*

*D) **I and II*

*E) **II and III *

Okay, so let’s list out the elements in this set. We know that 1 is in the set. If x= 1, then x – 3 = -2. So -2 is in the set. If x = -2 is in the set, then x – 3 = -5. So -5 is in the set.

By this point, the pattern should be clear: each term is three less than the previous term, giving us a sequence that looks like this: 1, -2, -5, -8, -11….

So we look at our options, and see we that only III is true. And we’re done. That’s it. The answer is C.

*Sure, Dave*, you may say. *That is much easier than any question I’m going to see on the GMAT. * First, this is an official question, so I’m not sure where you’re getting the idea that you’d never see a question like this. Second, you’d be surprised by how many test-takers get this wrong.

There is the temptation to assume that if 1 is in the set, then 4 must also be in the set. And note that this is, in fact, a possibility. If x = 4, then x – 3 = 1. But the question asks us what “must be” in the set. So it’s possible that 4 is in our set. But it’s also possible our set begins with 1, in which case 4 would not be included. This little wrinkle is enough to generate a substantial number of incorrect responses.

*Still, surely the questions get harder than this.* Well, yes. They do. *So what are you waiting for? *I’m not sure where this testy impatience is coming from, but if you insist:

*The sequence a _{1}, a_{2}, a_{3}, . . , a_{n} of n integers is such that a_{k} = k if k is odd and a_{k} = -a_{k-1} if k is even. Is the sum of the terms in the sequence positive?*

*1) **n is odd*

*2) **a _{n} is positive*

*Yikes!* Hey, you asked for a harder one. This question looks far more complicated than the previous one, but we can attack it the same way. Let’s establish our sequence:

*a _{1} *is the first term in the sequence. We’re told that

*a*. Well, 1 is odd, so now we know that

_{k}= k if k is odd*a*= 1. So far so good.

_{1}*a _{2} *is the second term in the sequence. We’re told that

*a*. 2 is even, so

_{k}= -a_{k-1}if k is even*a*Well, we know that

_{2}= -a_{2-1 }, meaning that a_{2}= -a_{1}.*a*= 1, so if

_{1}*a*then

_{2}= -a_{1 }*a*

_{2}= -1.So, here’s our sequence so far: 1, -1…

Let’s keep going.

a* _{3}* is the third term in the sequence. Remember that

*a*. 3 is odd, so now we know that

_{k}= k if k is odd*a*= 3.

_{3}*a _{4} *is the fourth term in the sequence. Remember that

*a*. 4 is even, so

_{k}= -a_{k-1}if k is even*a*We know that

_{4}= -a_{4-1 }, meaning that a_{4}= -a_{3}.*a*= 3, so if

_{3}*a*then

_{4}= -a_{3 }*a*

_{4}= -3.Now our sequence looks like this: 1, -1, 3, -3…

By this point we should see the pattern. Every odd term is a positive number that is dictated by its place in the sequence (the first term = 1, the third term = 3, etc.) and every even term is simply the previous term multiplied by -1.

We’re asked about the sum:

After one term, we have 1.

After two terms, we have 1 + (-1) = 0.

After three terms, we have 1 + (-1) + 3 = 3.

After four terms, we have 1 + (-1) + 3 + (-3) = 0.

Notice the trend: after every odd term, the sum is positive. After every even term, the sum is 0.

So the initial question, “*Is the sum of the terms in the sequence positive?” c*an be rephrased as,* “Are there an ODD number of terms in the sequence?”*

Now to the statements. Statement 1 tells us that there are an odd number of terms in the sequence. That clearly answers our rephrased question, because if there are an odd number of terms, the sum will be positive. This is sufficient.

Statement 2 tells us that *a _{n} is positive. a_{n} is *the last term in the sequence. If that term is positive, then, according to the pattern we’ve established, that term must be odd, meaning that the sum of the sequence is positive. This is also sufficient. And the answer is D, either statement alone is sufficient to answer the question.

Takeaway: sequence questions are nothing to fear. Like everything else on the GMAT, the main obstacle we need to overcome is the self-fulfilling prophesy that we don’t know how to proceed, when, in fact, all we need to do is simplify things a bit.

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*By David Goldstein, a Veritas Prep GMAT instructor based in Boston. You can find more articles written by him here.*