Repeated Question Patterns
Whether straightforward or difficult, almost every problem on GRE Quantitative Reasoning follows established templates, variations on timetested scenarios. In this post, we will define sequences and series, explain how they work, and then attempt an example problem.
Key to success on the GRE is the ability to deal with problems in multiple ways, both conventional and unconventional. This way you will get stuck less frequently and always have additional approaches to fall back on in the event one approach doesn't work out the way you want it to. Watch these multiple approaches in action with our example problem below.
Definitions
Let's begin with a couple definitions:

An arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant.
 e.g. a_{n} = 1 + (n1) for a sequence of the first n positive consecutive integers; if n = 100, then the sequence would comprise the numbers {1, 2, 3, ..., 98, 99, 100}
 What would the fiftieth term be? a_{50} = 1 + (50  1)= 1 + 49 = 50. The fiftieth term is 50.
 e.g. a_{n} = 2 + (n1)2 for a sequence of the first n positive consecutive even integers; if n = 50, then the sequence would comprise the numbers {2, 4, 6, ..., 96, 98, 100}. Note here that n denotes the number of members in the sequence, not the final number in the sequence.
 What would the twentieth term be? a_{20} = 2 + (20  1)2 = 2 + 19 ∙ 2 = 2 + 38 = 40. The twentieth term is 40.
 e.g. a_{n} = 1 + (n1) for a sequence of the first n positive consecutive integers; if n = 100, then the sequence would comprise the numbers {1, 2, 3, ..., 98, 99, 100}

An arithmetic series is the sum of the members of a finite arithmetic sequence.
 Consider our second arithmetic sequence above (a_{n} = 2 + (n1)2) and let n = 50. How could we find the sum of all the terms in the sequence {2, 4, 6, ..., 96, 98, 100}? We obviously don't want to start adding 2 + 4 + 6 + 8 etc. because that would take too long! Fortunately, something kinda cool happens.
 Let's start listing out each number in the sequence using its formula a_{n} = 2 + (n1)2, using 1, 2, 3 etc. for n all the way up to 50:
 (2 + (11)2) + (2 + (21)2) + ... + (2 + (491)2) + (2 + (501)2)
 (2) + (2 + 2) + ... + (2 + 96) + (2 + 98)
 2 + 4 + ... + 98 + 100
 Or we could express the same series thus:
 (100  (501)2) + (100  (502)2) + ... + (100  (5049)2) + (100  (5050)2)
 (100  98) + (100  96) + ... + (100  2) + (100)
 2 + 4 + ... + 98 + 100
 Look what happens when you add the first way of writing the series ((2 + (11)2) + (2 + (21)2) + ... + (2 + (491)2) + (2 + (501)2)) to the second way of writing the series (100  (501)2) + (100  (502)2) + ... + (100  (5049)2) + (100  (5050)2).
 For each of the 50 members of the sequence above, every addition in the first way of writing the series matches a subtraction in the second way of writing the series.
 (2 + (11)2) and (100  (5050)2)
 (2 + (21)2) and (100  (5049)2)
 (2 + (491)2) and (100  (502)2)
 (2 + (501)2) and (100  (501)2)
 So when we combine these two ways of writing the series, the additions and subtractions all cancel out, and we're left with (2 + 100) + (2 + 100) + ... + (2 + 100) + (2 + 100) fifty times:
 50(2 + 100)
 However, this is what happens when we add the series to itself. Now we have 2 times the series. We must divide this sum by 2 to get the series by itself:
 (50(2 + 100)) / 2
 More generally, this translates into the following equation for finding the sum of all the members of a sequence:

S_{n} = (n(a_{1} + a_{n})) / 2 where n is the number of members of the sequence, a_{1} is the first member of the sequence, and a_{n }is the nth member of the sequence.

For the GRE, you are not expected to memorize this equation. On GRE problems, sometimes an equation will be provided, as in the example below. However, it is helpful to understand why the equation works and to be familiar with how to use it; you'll find sequence and series problems far less intimidating. Now let's examine a sample series problem akin to what you might expect on the GRE.
Example
Consider the following example:
For any positive integer n, the sum of the first n positive integers equals (n(n + 1)) / 2. What is the sum of all the even integers between 49 and 101?
A. 1875
B. 1950
C. 2550
D. 4950
E. 5000
The equation given here is a simplified version of the more complete expression of the series equation given above. Unfortunately, this simplified version of the equation actually prevents us from solving the series directly!
If we do remember and use the complete equation, we can solve this problem quickly and directly.
Using the equation S_{n} = (n(a_{1} + a_{n})) / 2 to solve this problem, we would set a_{1 }= 50 and a_{n }= 100. Next we need to determine what n equals. How many even integers are between 49 and 101? Let's start listing out the values:
50 52 54 56 58
60 62 64 66 68
70 ...
80 ...
90 ...
100
For 50 through 98, we have 5 members each time. Thus, 5 x 5 = 25. 100 is by itself, so we have 26 members total. Thus n = 26. Using these values in the equation S_{n} = (n(a_{1} + a_{n})) / 2, we would have S_{n} = (26(50 + 100)) / 2, which gives us 1950, the correct answer.
What do we do if we don't remember the complete equation and must use the incomplete version provided?
Let's use the simplified (and less useful) series formula (n(n + 1)) / 2. Here we assume that a_{1 }= 1 and a_{n }= n. How could we solve this problem using this inferior formula?
Let's use some reasoning. Note that the even integers 2 through 100 represent the first 50 even integers. Thus, if we assumed that n = 50, we could plug 50 into our formula (n(n + 1)) / 2 to get (50 x 51) / 2. This would give us the sum of the sequence from 1 to 50. However, since we need the sum of the even integers, this series is only half the sum of the first 50 even integers, so we can double the sum! If you take (2(50 x 51)) / 2, the 2s cancel and you're left with 50 x 51 = 2550, but we're not done. We now have the sum of the first 50 even integers, but we're only interested in 50 through 100. In other words, we need to get rid of the sum of 2 through 48. How do we do this? Subtract! Find the sum of 2 through 48 using the same method as above. 2 through 48 are the first 24 even integers. In our formula, we'd get (2(24 x 25)) / 2, or 24 x 25 = 600. Now subtract this second series from the first. 2550  600 = 1950, the correct answer.
This method is likely the approach ETS expects students to use.
But there's still an even easier way!
If we're asked to calculate the sum of the even integers between 49 and 101, note that working from the smallest term up and the largest term down, every pair adds up to 150 (50 + 100, 52 + 98, 54 + 96, etc.). How many pairs do we have? Half of 26. 13. Indeed it works. Take 13 x 150 and you get the correct answer, 1950.
And there's another easy way!
You might also note that the average of 100 and 50 is 75. If instead of the sequence of the 26 even integers between 49 and 101, we just had the average 75 26 times, we could just take 75 x 26 to again get the correct answer, 1950.
Keep Going! Don't Get Stuck!
Four ways to do one problem!
 Direct approach using the complete series formula.
 Simplified but less direct approach using the formula the question provides.
 Alternative approach from noticing patterns by listing out numbers.
 Additional alternative approach from noticing the average of all our terms.
Why did we go through all that trouble above? Why didn't we just introduce the easy solution first? The answer to this question is important!
For every Quantitative problem on the GRE, there are multiple valid approaches to find the solution. In your preparation, you must investigate multiple approaches to the problems. As you practice and review, consider doing the same problem two or three different ways. In this practice, observe patterns in the questions, learn the "style" of the GRE, and determine which approaches work best for you. The best learning occurs when you discover for yourself different ways to solve problems. Even if an answer explanation provides one way possible solution, that's not the end of the story. There are always other approaches.
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