## Definitions

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

**, 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.**

*sequences*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

The number of terms in a sequence is ** inclusive**. The count must include both the first and last terms.

## Beware of “inclusive questions”

When the GRE says “inclusive” it means that they are counting the first and the last term of a given sequence. Sometimes the GRE doesn’t even say “inclusive,” it is just implied in the question.

## Example

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** (last – first) + 1, or L – F + 1**.

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

## Example

How many integers are there from 2 to 6?

### 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.

## 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 GRE 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 (last + first)/2, or (

*L*+

*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.

## Example

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

### Solution

- For this sequence,

(L + F)/2 = (10 + 25)/2

= 35/2 = 17.5. - You can easily check this answer:

(10 + 15 + 20 + 25)/4

= 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}

## Example

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

### Solution

There are (*L* – *F* + 1) = 6 – 2 + 1

= 5 items.

The mean of the numbers is

(last + first)/2 = (*L* + *F*)/2

= (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.

## Example

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

### Solution

The number of items is *L* – *F* + 1 = 100 – 1 + 1 = 100.

The mean of the sequence is

(*L* + *F*)/2 = (100 + 1)/2

= (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.*

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*.

## Example

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

### 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) = 6*x* + 30

6*x* + 30 = 3(2*x* + 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

## Example (

advanced)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 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 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