* 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 (
*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.

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.

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.

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. Use 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.

## 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)((L+F)/2)

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.

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.

### 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*.

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) = 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

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

### Solution

- All three terms are even integers so each has 2 as a factor. 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.
- 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 when x is divided by 3 can be 0, 1 or 2. If it 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 factors is divisible by 3.
- All the products of 3 with the factors of 2 (2, 4, and 8) will also be factors. Those factors will be 3(2) = 6, 3(4) = 12, and 3(8) = 24. The list of factors includes 2, 3, 4, 6, 8, 12, and 24.