AÂ ** sequence**Â is an ordered list of numbers. Each number contained in a sequence is called aÂ

**.**

*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 position-number of a term.

*a _{n}*Â represents the value of theÂ

*n*

^{th}Â term.

*d*Â is the difference between terms.

*r*Â 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 |

*a*=

_{n}*n*+ 4

### Solution

Substitute the term number for *n* in the equation that describes the sequence.

The first term *a*_{1} would be *a*_{1} Â = 1 + 4 = 5.

The second term *a*_{2} would be *a*_{2}Â = 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 Ã— 5.

The second term *a*_{2 }= 10 = 2 Ã— 5.

The third term *a*_{3} = 15 = 3 Ã— 5.

The fourth term *a*_{4} = 20 = 4 Ã— 5.

So the rule for the sequence is that the *n*th term is *n* Ã— 5.

*a _{n}* = 5

*n*

## Arithmetic Sequence

In anÂ ** arithmetic sequence**Â 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,Â

*a*Â =Â

_{n}*n*Â + 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*Â Â Â Â *a _{n}*Â =Â

*a*

_{1Â }+ (

*n*Â â€“ 1)

*d*

^{th}term in this sequence:Â 2, 5, 8, 11, 14, …

### Solution

Sometimes the easiest way to find a term is to just do the calculations.

**term**Â Â Â Â Â Â *a*_{1}Â Â Â Â Â Â *a*_{2}Â Â Â Â Â Â *a*_{3}Â Â Â Â Â Â *a*_{4}Â Â Â Â Â Â *a*_{5}Â Â Â Â Â Â *a*_{6}Â Â Â Â Â Â *a*_{7} Â Â Â Â Â *Â a*_{8}Â Â Â Â Â Â *a*_{9 }**value **Â Â Â Â Â Â 2Â Â Â Â 5Â Â Â Â Â 8Â Â Â Â 11Â Â Â Â 14Â Â Â Â **17Â Â Â Â Â Â 20Â Â Â Â Â Â 23Â Â Â 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 â€“ 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.

___Â Â ___Â Â ___Â Â ___Â Â __ 19 __Â Â ___

*a*_{1}Â Â Â *a*_{2}Â Â Â *a*_{3}Â Â Â *a*_{4}Â Â Â *a*_{5}Â Â Â Â *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} â€“ 3 = *a*_{5} = 19, and *a*_{4} = 22.

Continue to fill in the blanks.

__ 31 __Â __ 28 __Â __ 25 __Â __ 22 __Â __ 19__Â ___

*a*_{1}Â Â Â *a*_{2}Â Â Â *a*_{3}Â Â Â *a*_{4}Â Â Â *a*_{5}Â Â Â *a*_{6}

So the second term *a*_{2} = 28. The correct answer choice is (C).

## Arithmetic Series

AÂ ** series**Â is the sum of the terms in a sequence.

*sequence*:Â Â 5, 10, 15, 20

*series*:Â Â Â Â Â Â Â Â Â 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 withÂ *a*_{1}Â = 2 andÂ *d*Â = 3.

*a*_{100}Â = 2 + (100 â€“ 1)(3) = 2 + (99)(3) = 299 Â Â Â Â The 100^{th}Â term is 299.

So the first term is 2, the 100^{th}Â 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 aÂ ** geometric sequence**Â 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,Â

*a*Â = 5

_{n}*n*Â 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*Â Â Â Â Â *a _{n}*Â =Â

*a*

_{1}Â Ã—Â

*rÂ*

^{n}^{Â â€“ 1}

^{th}term in the sequence:Â 3, 6, 12, 24, …

### Solution

First find the ratio between terms.

\dfrac{6}{3} = 2

\dfrac{12}{6} = 2

\dfrac{24}{12} = 2

UseÂ *r*Â = 2 andÂ *a*_{1}Â = 3 in the formula.

*a*_{10} = 3 Ã— 2^{(10 – }^{1)}Â

= 3 Ã— 2^{9}Â = 3 Ã— 512 = 1,536Â Â Â Â Â Â Â Â Â Â Â

The 10^{th}Â term is 1,536.

^{th}Â and 8

^{th}terms in the sequence:Â -729, 243, -81, 27, …

### 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 useÂ *a*_{1}Â = -729 in the formula and look for the 7^{th}Â and 8^{th}Â terms. But if you useÂ *a*_{4}Â = 27 as the first term, the calculations will be easier. You will look for the fourth and fifth terms after 27.

*a*_{4}Â = 27Â *Ã—* (-\,\dfrac{1}{3})^{4â€“1}Â = 27Â *Ã—* (-\,\dfrac{1}{3})^{3}Â

= 27Â *Ã—* (-\,\dfrac{1}{27}) = \dfrac{27}{-\,27} = -1

For (-\,\dfrac{1}{3})^{3} rememberÂ *negativeÂ Ã—Â negativeÂ Ã—Â negative = negative.*

*a*_{5}Â = 27Â *Ã—* (-\,\dfrac{1}{3})^{5â€“1}Â = 27Â *Ã—* (-\,\dfrac{1}{3})^{4}Â

= 27Â *Ã—* (\dfrac{1}{81}) = \dfrac{27}{81} = \dfrac{1}{3}

For (-\,\dfrac{1}{3})^{4}Â rememberÂ *negativeÂ Ã—Â 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 __Â __ 243 __Â __ -81 __Â __ 27 __Â __ -9 __Â __ 3 __Â __ -1 __Â \,\underline{\dfrac{1}{3}}\,

*a*_{1}Â Â Â Â Â Â Â *a*_{2}Â Â Â Â Â *a*_{3}Â Â Â *a*_{4}Â Â Â *a*_{5}Â Â Â *a*_{6}Â Â Â *a*_{7}Â Â Â *a*_{8}

So the 7^{th}Â 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 Ã— 4 = -8

*a*_{6} = 4 Ã— -8 = -32

*a*_{7} = -8 Ã— -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,Â **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).

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## Definitions

** Consecutive integers**Â are integers that follow one another with a difference of 1. There are other types of consecutive patterns orÂ

**, such as even integers where the difference between consecutive even integers is 2.Â Two common question types about sequences of consecutive integers are the number of items in the sequence and the sum of the sequence.**

*sequences*4, 6, 8, 10 Â Â Â Â Â Â

difference = 2: consecutive even integers

3, 5, 7, 9, 11, 13 Â

difference = 2: consecutive odd integers

3, 5, 7, 11, 13 Â Â Â

difference varies: consecutive prime numbers

10, 15, 20, 25 Â Â Â

difference = 5: multiples of 5

## Number of Integers

Sometimes a question might describe an inclusive** sequence**.Â This means the count must include both the first and last terms.

**A fence consists of fence posts that are 1 foot wide and bridges that are 9 feet long connecting each post. How many fence posts would be required for a 100 foot fence?**

### 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**Â (last â€“ first) + 1, orÂ LÂ â€“Â FÂ + 1**.

Note that this formula only works for sequences with a difference of 1 between terms.

**How many integers are there from 2 to 6?**

### Solution

You might think you just subtract.

Last â€“ first = 6 â€“ 2 = 4.

But try counting the numbers from 2 to 6.Â **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 theÂ *arithmetic mean*Â of the sequence.Â (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.Â 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 willÂ *not* be an integer if there is an even number of terms.

**What is the mean of the sequence 10, 15, 20, 25?**

### 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}Â

= \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}

**What is the sum of the integers from 2 to 6?**

### Solution

There are (*L*Â â€“Â *F* + 1) = 6 â€“ 2 + 1

= 5 items.

The mean of the numbers is

\dfrac{(last + first)}{2} = \dfrac{(\textit{L}Â + \textit{F})}{2}

= \dfrac{(6 + 2)}{2} = 4.

The sum of the numbers is therefore 5 *Ã—* 4 = 20.

You can easily check this answer:

2 + 3 + 4 + 5 + 6 = 20.

**What is the sum of the integers from 1 to 100?**

### Solution

The number of items isÂ *L*Â â€“Â *F*Â + 1 = 100 â€“ 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 *Ã—*Â 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 isÂ *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 isÂ *even*.

**Is the sum of six consecutive even integers divisible by 3?**

### Solution

Write the sequence using variables.Â LetÂ *x*Â be an even integer, and the difference between even integers is 2.

*x*Â + (*x*Â + 2) + (*x*Â + 4) + (*x*Â + 6) + (*x*Â + 8)

+ (*x*Â + 10) = 6*x*Â + 30

6*x*Â + 30 = 3(2*x*Â + 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 anÂ *even* number of integers in a consecutive integer sequence, the product will always be divisible by 2.

Overall, the product ofÂ *n*Â consecutive integers is divisible byÂ *n*!

*n*! is “*n*Â factorial.”Â For a positive integerÂ *n*,Â *n*! is the product of all positive integers less than or equal toÂ *n*.

Example:

6! = 6Â *Ã—*Â 5Â *Ã—*Â 4Â *Ã—*Â 3Â *Ã—*Â 2Â *Ã—*Â 1 = 720

**List 6 factors of the product of any 3 consecutive even integers.**

### 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 2Â *Ã—*Â 2Â *Ã—*Â 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 2*x*, where *x* is some integer. Then the sequence is

2*x*, 2*x* + 2, 2*x* + 4

or 2*x*, 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

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