A sequence is an ordered list of numbers. Each number contained in a sequence is called a term.
A sequence is defined by an equation or rule. The rule can be applied to each term of a sequence to generate the next term in the sequence. The ellipsis (…) at the end of a sequence means the sequence continues using the same rule.
The key to solving sequence problems is to determine the relationship between the terms.
What is the next term in the sequence?
2, 4, 6, 8, …
Solution
2, 4, 6, 8, …
Here you can probably assume the next number of the sequence is 10, but always be careful with snap assumptions on sequence problems.
What is the next term in the sequence?
1, 2, 3, 5, …
Solution
1, 2, 3, 5, …
Here you cannot be sure of the next term. If the sequence is odd integers, the next term would be 9. If the rule is each term is the sum of the previous 2 terms, the next term could be 8.
Don’t Assume
The GMAT likes to trick you with a sequence like:
What is the next term in this sequence? 3, 5, 7, ____, …
You may immediately think “consecutive odd numbers” and assume the next term is 9. But it could be “consecutive primes” and the next term would be 11. On Data Sufficiency questions in particular, be careful not to make assumptions about the rule in a sequence.
Writing the Rules
The key to solving sequence problems is to determine the relationship between the terms in the sequence. The rule can be applied to each term of a sequence to generate the next term in that sequence.
There is a common set of variables used in the equation or rule for a sequence.
Subscripts are used to give the positionnumber of a term.
a_{n} represents the value of the n^{th} term.
d is the difference between terms.
r is the ratio between terms.
So for the sequence 2, 4, 6, 8, 10:
value
term & number
pattern
2
a_{1} = 2
a_{1} = 2
4
a_{2} = 4
a_{2} = a_{1} + 2 = 4
6
a_{3}= 6
a_{3} = a_{2} + 2 = 6
8
a_{4} = 8
a_{4} = a_{3} + 2 = 8
10
a_{5} = 10
a_{5} = a_{4} + 2 = 10
value  term & number  pattern 
2  a_{1} = 2  a_{1} = 2 
4  a_{2} = 4  a_{2} = a_{1} + 2 = 4 
6  a_{3}= 6  a_{3} = a_{2} + 2 = 6 
8  a_{4} = 8  a_{4} = a_{3} + 2 = 8 
10  a_{5} = 10  a_{5} = a_{4} + 2 = 10 
Solution
Substitute the term number for n in the equation that describes the sequence.
The first term a_{1} would be
a_{1} = 1 + 4 = 5.
The second term a_{2} would be
a_{2} = 2 + 4 = 6.
The third term a_{3 }would be
a_{3} = 3 + 4 = 7.
The fourth term a_{4 }would be
a_{4} = 4 + 4 = 8.
Solution
Look for a pattern that uses the position number n.
The first term a_{1} = 5 = 1 × 5.
The second term a_{2 }= 10 = 2 × 5.
The third term a_{3} = 15 = 3 × 5.
The fourth term a_{4} = 20 = 4 × 5.
So the rule for the sequence is that the nth term is n × 5.
a_{n} = 5n
Arithmetic Sequence
In an arithmetic sequence the difference between consecutive terms is constant. You add (or subtract) the same amount to go from one term to the next. From the Example above, a_{n} = n + 4 is an arithmetic sequence.
There is a general form for the formula of an arithmetic sequence. This formula allows you to calculate the value of any term in an arithmetic sequence with just the value of the first term and the difference.
arithmetic sequence a_{n} = a_{1 }+ (n – 1)d
Solution
Sometimes the easiest way to find a term is to just do the calculations.
term a_{1} a_{2} a_{3} a_{4} a_{5} a_{6} a_{7} a_{8} a_{9 }value 2 5 8 11 14 17 20 23 26
term  value 
a_{1}  2 
a_{2}  5 
a_{3}  8 
a_{4}  11 
a_{5}  14 
a_{6}  17 
a_{7}  20 
a_{8}  23 
a_{9}  26 
But for 100 terms, there has to be a better way.
From looking at the first 5 terms, you can see the difference is 3. Use the formula for arithmetic sequence with a_{1}= 2 and d = 3.
a_{100} = 2 + (100 – 1)(3) = 2 + (99)(3) = 299
The 100^{th} term is 299.
(B) 13
(C) 28
(D) 30
(E) 31
Solution
A good technique to visualize this short sequence is to use blanks.
___ ___ ___ ___ 19 ___
a_{1} a_{2} a_{3} a_{4} a_{5} a_{6}
You are told that each term is 3 less than the term immediately preceding it.
Does that mean that the fourth term, the one immediately preceding 19, will be 3 more or 3 less? Be sure to read carefully!
19 is 3 less than the term preceding it. So a_{4} – 3 = a_{5} = 19, and a_{4} = 22.
Continue to fill in the blanks.
31 28 25 22 19 ___
a_{1} a_{2} a_{3} a_{4} a_{5} a_{6}
So the second term a_{2} = 28. The correct answer choice is (C).
Arithmetic Series
A series is the sum of the terms in a sequence.
sequence: 5, 10, 15, 20
series: 5 + 10 + 15 + 20 = 50
There is a formula for arithmetic series. The formula allows you to calculate the sum of the arithmetic sequence using the number of terms and the values of the first and last terms.
The sum of an arithmetic sequence is the mean of the first and last terms multiplied by the number of terms.
arithmetic series
S_{n} = n \Big( \dfrac{\textit{a}_{\displaystyle{1}} + \textit{a}_{\displaystyle{\textit{n}}}}{2}\Big)
Notice how similar this is to the formula for adding consecutive integers, such as the sum of all the integers from 1 to 100. The concept here is the same: take the mean of the terms and multiply it by the number of terms.
(Note: This is covered further here: Consecutive Numbers.)
2 + 5 + 8 + 11 + 14 + …
Solution
With this many terms, there must be a shortcut.
The first step is to calculate the value of the 100th term. An Example above does this calculation:
Use the formula for arithmetic sequence with a_{1} = 2 and d = 3.
a_{100} = 2 + (100 – 1)(3) = 2 + (99)(3) = 299 The 100^{th} term is 299.
So the first term is 2, the 100^{th} term is 299, and there are 100 terms.
S_{100} = 100 \Big( \dfrac{2 + 299}{2}\Big) = 100 \Big( \dfrac{301}{2}\Big) = 100(150.5) = 15,050
Geometric Sequence
In a geometric sequence the ratio between consecutive terms is constant. You multiply by the same number to go from one term to the next. From an Example above, a_{n} = 5n is a geometric sequence where the ratio is 5.
There is a formula for a geometric sequence that allows you to calculate the value of any term in a geometric sequence with just the value of the first term and the ratio.
geometric sequence a_{n} = a_{1} × r ^{n}^{ – 1}
Solution
First find the ratio between terms.
\dfrac{6}{3} = 2
\dfrac{12}{6} = 2
\dfrac{24}{12} = 2
Use r = 2 and a_{1} = 3 in the formula.
a_{10} = 3 × 2^{(10 – }^{1)}
= 3 × 2^{9} = 3 × 512 = 1,536
The 10^{th} term is 1,536.
Solution
First find the ratio between terms.
\dfrac{243}{729} = \,\dfrac{1}{3}
\\[2ex]\dfrac{81}{243} = \,\dfrac{1}{3}
\\[2ex]\dfrac{27}{81} = \,\dfrac{1}{3}\\[2ex]
So r = \,\dfrac{1}{3}.
Method 1
You could use a_{1} = 729 in the formula and look for the 7^{th} and 8^{th} terms. But if you use a_{4} = 27 as the first term, the calculations will be easier. You will look for the fourth and fifth terms after 27.
a_{4} = 27 × (\,\dfrac{1}{3})^{4–1} = 27 × (\,\dfrac{1}{3})^{3}
= 27 × (\,\dfrac{1}{27}) = \dfrac{27}{\,27} = 1
For (\,\dfrac{1}{3})^{3} remember
negative × negative × negative
= negative.
a_{5} = 27 × (\,\dfrac{1}{3})^{5–1} = 27 × (\,\dfrac{1}{3})^{4}
= 27 × (\dfrac{1}{81}) = \dfrac{27}{81} = \dfrac{1}{3}
For (\,\dfrac{1}{3})^{4} remember
negative × negative = positive.
Method 2
Another technique for this short sequence is to use blanks. This makes it easier to see the alternating positive and negative terms.
729 243 81 27 9 3 1 \,\underline{\dfrac{1}{3}}\,
a_{1} a_{2} a_{3} a_{4} a_{5} a_{6} a_{7} a_{8}
So the 7^{th} and 8^{th} terms are 1 and \dfrac{1}{3}.
(B) 32
(C) 32
(D) 256
(E) 256
Solution
Method 1
Rather than a ratio between terms, there is a rule. Though you could write the rule for this short sequence, it is easier to use blanks or a list.
a_{1} = 1
a_{2} = 2
a_{3} = 2
a_{4} = 4
a_{5} = 2 × 4 = 8
a_{6} = 4 × 8 = 32
a_{7} = 8 × 32 = 256
The correct answer choice is (D).
Method 2
Use the strategy of reading the answers first. You can see that there are two basic pieces of information you need to find: whether the number is positive or negative, and how big it is.
First find the pattern for positive and negative without doing the calculations:
positive, negative, negative, positive, negative, negative, positive.
If you quickly calculate the fifth term, just looking at the change between the fourth term and the fifth terms shows you the value increases quickly. From that, you can infer that the largest value will be correct. The largest positive value is 256, which is (D).
Video Quiz
Sequence
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100 seconds per question
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Before attempting this problem, be sure to review this section on data sufficiency questions.
Definitions
Consecutive integers are integers that follow one another with a difference of 1. There are other types of consecutive patterns or sequences, such as even integers where the difference between consecutive even integers is 2. Two common question types about sequences of consecutive integers are the number of items in the sequence and the sum of the sequence.
4, 6, 8, 10
difference = 2: consecutive even integers
3, 5, 7, 9, 11, 13
difference = 2: consecutive odd integers
3, 5, 7, 11, 13
difference varies: consecutive prime numbers
10, 15, 20, 25
difference = 5: multiples of 5
Number of Integers
Sometimes a question might describe an inclusive sequence. This means the count must include both the first and last terms.
Solution
A fence post plus a bridge covers 10 feet, so it takes 10 of those pairs to cover 100 feet. The trick here is to realize that there is also a post at the end. This means that there would be 10 posts plus the last post, for a total of 11 posts.
The formula for the number of consecutive integers in a sequence is (last – first) + 1, or L – F + 1.
Note that this formula only works for sequences with a difference of 1 between terms.
Solution
You might think you just subtract.
Last – first = 6 – 2 = 4.
But try counting the numbers from 2 to 6. 2, 3, 4, 5, 6. The answer is 5. Using the formula, 6 – 2 + 1 = 5.
Beware of “inclusive questions”
When the GMAT says “inclusive” it means that they are counting the first and the last term of a given sequence. Sometimes the GMAT doesn’t even say “inclusive,” it is just implied in the question.
Sum of a Sequence
What is the sum of the numbers from 1 to 100?
Adding all the integers would take too much time for a GMAT problem, so there must be a formula. The formula has two pieces: the number of integers and the arithmetic mean of the sequence. (Arithmetic mean will be covered in Chapter 7.)
To find the mean of a sequence where the difference between items is constant, you don’t need to add all the items. The mean of the sequence is \dfrac{(last + first)}{2}, or \dfrac{(\textit{L} + \textit{F})}{2}.
The mean of a sequence of consecutive integers will be an integer if there is an odd number of terms.
The mean will not be an integer if there is an even number of terms.
Solution

 For this sequence,
\dfrac{(\textit{L} + \textit{F})}{2} = \dfrac{(10 + 25)}{2}
= \dfrac{35}{2} = 17.5.
 For this sequence,
 You can easily check this answer:
\dfrac{(10 + 15 + 20 + 25)}{4}
= \dfrac{70}{4} = 17.5.
The sum of a sequence with a constant difference of 1 is the number of terms in the sequence times the mean of the sequence, or: (L – F + 1) \dfrac{\,\,(\textit{L}+\textit{F})\,\,}{2}
Solution
There are (L – F + 1) = 6 – 2 + 1
= 5 items.
The mean of the numbers is
\dfrac{(last + first)}{2} = \dfrac{(\textit{L} + \textit{F})}{2}
= \dfrac{(6 + 2)}{2} = 4.
The sum of the numbers is therefore 5 × 4 = 20.
You can easily check this answer:
2 + 3 + 4 + 5 + 6 = 20.
Solution
The number of items is
L – F + 1 = 100 – 1 + 1 = 100.
The mean of the sequence is
\dfrac{(\textit{L} + \textit{F})}{2} = \dfrac{(100 + 1)}{2}
= \dfrac{(101)}{2} = 50.5.
So the sum of the numbers is
100 × 50.5 = 5050.
How to Solve Summation Questions
Divisibility and Sums of Consecutive Integers
(NOTE: This is an advanced topic. The general topic of divisibility is covered in the next section.
Rule: If the number of terms in a sequence is odd, the sum will be a multiple of the number of terms.
There are 5 numbers in this set of consecutive integers:
1 + 2 + 3 + 4 + 5 = 15.
YES, 15 is a multiple of 5. The number of terms is odd.
There are 6 numbers in a set of consecutive integers:
11 + 12 + 13 + 14 + 15 + 16 = 81.
NO, 81 is not a multiple of 6. The number of terms is even.
Solution
Write the sequence using variables. Let x be an even integer, and the difference between even integers is 2.
x + (x + 2) + (x + 4) + (x + 6) + (x + 8)
+ (x + 10) = 6x + 30
6x + 30 = 3(2x + 10), so the sum is divisible by 3.
Divisibility of Products of Consecutive Integers
In a sequence of consecutive numbers, the product will always be divisible by the number of integers.
If there is an even number of integers in a consecutive integer sequence, the product will always be divisible by 2.
Overall, the product of n consecutive integers is divisible by n!
n! is “n factorial.” For a positive integer n, n! is the product of all positive integers less than or equal to n.
Example:
6! = 6 × 5 × 4 × 3 × 2 × 1 = 720
Solution
All three terms are even integers so each has 2 as a factor. Because there are three terms, their product will be divisible by 2 × 2 × 2. This means 2, 4, and 8 are all factors of their product.
At least one of the terms must be divisible by 3:
Look at any sequence of three even integers: 2, 4, 6 or 16, 18, 20. One term is always divisible by 3.
To see this algebraically, let’s denote the first even integer by 2x, where x is some integer. Then the sequence is
2x, 2x + 2, 2x + 4
or 2x, 2(x + 1), 2(x + 2).
Let’s take a look at x, x + 1, x + 2. The remainder of any number divided by 3 can be 0, 1 or 2. If the remainder is 0, then x is divisible by 3. If it is 1, then x + 2 is divisible by 3. If it is 2, then x + 1 is divisible by 3. So, in any case, one of the terms is divisible by 3.
Now we know that 2, 3, 4 and 8 are all factors of the product, but the question asks for 6 factors. We can simply multiply any of these factors by each other to get additional factors.
3(2) = 6
3(4) = 12
So the list of factors includes
2, 3, 4, 6, 8, 12
Video Quiz
Sequence
Best viewed in landscape mode
3 questions with video explanations
100 seconds per question
Are you sure you want to refresh the question?
https://www.youtube.com/watch?v=sx9jki181oc
https://www.youtube.com/watch?v=W54nXLFz0TU
https://www.youtube.com/watch?v=XGihGsWFeu0
Before attempting this problem, be sure to review this section on data sufficiency questions.