**Introduction**

Flipping through the pages of the book, on each page you can see the number of the page and it helps you know what is the number of the next page, how many pages are between 35th and 62th pages or how many pages you have read so far. Why is this happening? Because the pages are numbered with consecutive natural numbers and it makes your life easier. Sometimes objects are numbered consecutively in a certain pattern. For example, in a circus, each next row of a sector has 2 more seats than the previous one. Depending on whether the first row has an even number or an odd number of seats, we will get consecutive even or odd natural numbers of seats in this sector.

**What are consecutive numbers?**

The easiest way to get acquainted with consecutive numbers is to remind natural numbers. Natural numbers are numbers used in counting, 1, 2, 3, 4, 5, 6, … As we can see the difference between any two neighbor natural numbers is 1. Hence, consecutive natural numbers are the numbers that follow each other in order from the smallest to the greatest, increasing by 1.

So, we are ready to define consecutive numbers. Numbers that follow each other continuously in regular counting order or pattern are called consecutive numbers. Natural numbers, whole numbers, integers – all these numbers follow each other continuously in regular ascending order and are consecutive numbers.

In practice, we often start counting consecutive numbers from any number.

**EXAMPLE:** The years of birth of six children are six consecutive natural numbers, and the oldest child was born in 2007. List all the years of birth of all children.

**SOLUTION:** If the oldest child was born in 2007, the next child was born in 2008, and all other children – in 2009, 2010, 2011, 2012 years. Therefore, the list of the years of birth of all children is 2007, 2008, 2009, 2010, 2011, 2012.

Sometimes we solve real-life problems involving consecutive natural numbers in reverse order, from largest to smallest.

**EXAMPLE:** Today is Friday, 13^{th}. What day of the week and what number was six days ago?

**SOLUTION:** Today – Friday, 13^{th}

1 day before – Thursday, 12^{th}

2 days before – Wednesday, 11^{th}

3 days before – Tuesday, 10^{th}

4 days before – Monday, 9^{th}

5 days before – Sunday, 8^{th}

6 days before – Saturday, 7^{th}

If we add 1 to any natural number, we get the next number called the successor of the given number. For example, the successor of 26 is 26+1=27. If we subtract 1 from any natural number, we get the previous number called the predecessor of the given number. So, number 43 is the predecessor of 44 because 44-1=43. In the sets of natural numbers and whole numbers each number has the successor. In the set of natural numbers, number 1 has no predecessor because 1-1=0 is not a natural number. In the set of whole numbers, number 1 has the predecessor but number 0 hasn’t. In the set of integers each number has both, the successor and predecessor.

**Predecessor of the number, the number itself and successor of the number are always three consecutive numbers.**

This can be written using a mathematical notation. If n is a given number, then its predecessor is the number n-1 and its successor is the number n+1. The list of these three numbers looks like

n-1, n, n+1

If we want to mathematically set consecutive numbers starting with an arbitrary number n, it is also easy to do, just write after each number its successor,

n, n+1, n+2, n+3, n+4, n+5, …

and get the list of consecutive numbers written in ascending order. Here number n+1 is the successor of the number n, number n+2 is the successor of the number n+1, and so on.

On the other hand, if we need consecutive numbers written in descending order, then we start with an arbitrary number n and write after each number its predecessor,

n, n-1, n-2, n-3, n-4, n-5, …

Here number n-1 is the predecessor of the number n, number n-2 is the predecessor of the number n-1, and so on.

**EXAMPLE: **Write seven consecutive integers:

a) in ascending order starting from number 163;

b) in descending order starting from number 96.

**SOLUTION:** a) Start with number 163. Its successor is 163+1=164, so the next number in the list is number 164. The successor of the number 164 is number 164+1=165. Continuing to calculate the successors of numbers, we obtain the list of the following seven numbers

163, 164, 165, 166, 167, 168, 169

b) Start with number 96. Its predecessor is 96-1=95, so the next number in the list is number 95. The predecessor of the number 95 is number 95-1=94. Continuing to calculate the predecessors of numbers, we obtain the list of the following seven numbers

96, 95, 94, 93, 92, 91, 90

So, we already know how to write consecutive numbers. Then the question arises how to count how many consecutive numbers are written.

- If we have ascending sequence of natural numbers, n, n+1, n+2, …,n+k, there are k+1 numbers in this sequence.
- If we have descending sequence of natural numbers, n, n-1, n-2, …,n-k, there are also k+1 numbers in this sequence.

Why is that? We add or subtract all consecutive numbers from 1 to k. There are k numbers from 1 to k and plus initial number will give us k+1 numbers in total.

**EXAMPLE:** How many natural numbers are from 39 to 71 inclusive?

**SOLUTION:** We can represent number 71 as the sum 39+32. If the greatest number which we add is 32, then there are 32+1=33 natural numbers from 39 to 71 inclusive.

**Consecutive odd and even numbers**

Sometimes we may not be interested in all consecutive numbers, but only in even consecutive numbers or in odd consecutive numbers. Suppose that geese graze on the lawn. Each goose has two legs and we can count the geese by heads 1, 2, 3, … or by legs 2, 4, 6, …

Even numbers that follow each other continuously in regular counting order or pattern are called consecutive even numbers, odd numbers that follow each other continuously in regular counting order or pattern are called consecutive odd numbers.

Example of even consecutive numbers: -12, -10, -8, -6, …

Example of odd consecutive integers: 19, 21, 23, 25, 27, …

The difference between two consecutive even numbers is always 2. If we want to mathematically set consecutive even numbers starting with an arbitrary even number n, write after each number the next number greater by 2 than the previous one,

n, n+2, n+4, n+6, n+8, …

The difference between two consecutive odd numbers is also 2. If we want to mathematically set consecutive odd numbers starting with an arbitrary odd number n, write after each number the next number greater by 2 than the previous one,

n, n+2, n+4, n+6, n+8, …

It is worth noting we do not distinguish between these two sequences, we need to know from which number, even or odd, each of these sequences starts. To avoid this, you can write consecutive even and consecutive odd numbers in another way.

Since all even numbers are evenly divisible by 2, the set of even natural numbers is usually written as

2, 4, 6, …, 2k-2, 2k, 2k+2, …

All remaining natural numbers are odd numbers, so the set of odd numbers is written as

1, 3, 5, …, 2k-1, 2k+1, …

Such notations are used not only for natural numbers, but for integers as well. For example, even integers are written as

…, -4, -2, 0, 2, 4, …, 2k, …

and odd integers are

…,-3, -1, 1, 3, …, 2k+1, …

These notations often help in solving a number of problems.

**EXAMPLE:** The sum of three consecutive odd numbers is 111. What are these numbers?

**SOLUTION:** Let 2k-1 be the smallest odd consecutive number, then the other two numbers are 2k-1+2=2k+1 and 2k+1+2=2k+3. The sum of these numbers equals to

(2k-1)+(2k+1)+(2k+3)=111

Solve this equation:

2k-1+2k+1+2k+3=111

6k+3=111

6k=111-3

6k=108

k=108÷6

k=18

Therefore, the smallest of the numbers is 2(18)-1=36-1=35,

the next number is 2k+1=2(18)+1=36+1=37,

and the greatest number is 2k+3=2(18)+3=36+3=39.

**Other sets of consecutive numbers**

In the previous topic, we described how we can represent even and odd numbers. But even numbers are the numbers that give the remainder of 0 when divided by 2. What if we consider division by 3 instead of 2? When divided by 3, three different remainders are possible, 0,1 and 2, so the integers that

- are evenly divisible by 3 are consecutive numbers from the list

…,-3, 0, 3, 6, 9, …, 3k, …

- leave the remainder of 1 when divided by 3 are consecutive numbers from the list

…,-5,-2, 1, 4, 7, …, 3k+1, …

- leave the remainder of 2 when divided by 3 are consecutive numbers from the list

…,-4,-1, 2, 5, 8, …, 3k+2, …

Thus, these are three patterns of consecutive integers that leave zero-remainder, remainder of 1 or 2 when divided by 3. In the same way, we can break the set of integers into m non-intersecting sequences of numbers by their remainders of division by m. It is said that each such sequence of consecutive numbers can be given by a general rule. For example, all numbers that give the remainder of 3 when divided by 5 can be given by a general rule 5k+3, where k is integer.

**EXAMPLE:** What numbers are represented by a general rule 7k-3, where k is integer? Can you represent these numbers using another rule?

**SOLUTION:** Write down the first few terms according to this rule:

- when k=1, 7k-3=7⋅1-3=4;
- when k=2, 7k-3=7⋅2-3=11;
- when k=3, 7k-3=7⋅3-3=18;
- when k=4, 7k-3=7⋅4-3=25;
- when k=5, 7k-3=7⋅5-3=32.

All these numbers leave the remainder of 4 when divided by 7. Therefore, the list of these consecutive numbers is

…,-10, -3, 4, 11, 18, 25, 32, …, 7k+4, …

By the way, recording all such consecutive numbers we managed to notice another rule that specifies numbers of this type: the numbers are of the form 7k+4, where k is integer.

**Properties of consecutive numbers**

**PROPERTY 1:** For any two consecutive numbers, the difference between each pair of two consecutive numbers is fixed. The difference between two consecutive integers is always 1. If you take two consecutive integers n and n+1, their difference is (n+1)-n=1.

**PROPERTY 2:** For any two consecutive numbers, the sum of these numbers is always odd. If you take two consecutive numbers, then one of them is necessarily even and the other – odd. The sum of even and odd numbers is always odd.

**PROPERTY 3:** For any two consecutive odd numbers, the difference is always 2. If the smaller of two consecutive odd numbers is 2k-1, then the greater odd number is 2k+1 and the difference of these numbers is (2k+1)-(2k-1)=2k+1-2k+1=2.

**PROPERTY 4:** For any two consecutive odd numbers, the sum of these numbers is always even. If the smaller of two consecutive odd numbers is 2k-1, then the greater odd number is 2k+1 and the sum of these numbers is (2k+1)+(2k-1)=2k+1+2k-1=4k that is an even number.

**PROPERTY 5:** For any two consecutive even numbers, the difference is always 2. If the smaller of two consecutive even numbers is 2k, then the greater even number is 2k+2 and the difference of these numbers is (2k+2)-(2k)=2k+2-2k=2.

**PROPERTY 6:** For any two consecutive even numbers, the sum of these numbers is always even. If the smaller of two consecutive even numbers is 2k, then the greater even number is 2k+2 and the sum of these numbers is (2k+2)+(2k)=2k+2+2k=4k+2=2(2k+1) that is an even number as a product of even and odd numbers.

**PROPERTY 7:** If n is an odd number, then the sum of n consecutive numbers will be evenly divisible by n.

**Patterns with consecutive numbers**

Very often we try to visualize mathematical facts. Actually, consecutive numbers are very good for such visualization. Here are some patterns that illustrate the relationships between consecutive numbers.

I. If we add two consecutive numbers together, the sum is an odd number,

1+2=3

2+3=5

3+4=7

4+5=9

5+6=11

n+(n+1)=2n+1

II. If we add any three consecutive numbers together it will always equal to a multiple of 3,

1+2+3=6

2+3+4=9

3+4+5=12

4+5+6=15

(n-1)+n+(n+1)=3n

III. If we add any five consecutive numbers together it will always equal to a multiple of 5,

1+2+3+4+5=15

2+3+4+5+6=20

3+4+5+6+7=25

4+5+6+7+8=30

(n-2)+(n-1)+n+(n+1)+(n+2)=5n

IV. If we add any odd number k of consecutive numbers together it will always equal to a multiple of k,

(n-$\frac{k-1}{2}$)+..+(n-1)+n+(n+1)+…+(n+$\frac{k-1}{2}$)=kn

V. The sum of n consecutive odd integers is equals to n^{2},

1^{2}=1

2^{2}=1+3

3^{2}=1+3+5

4^{2}=1+3+5+7

n^{2}=1+3+5+…+(2n-1)

VI. The sum of n odd integers starting after the previous series equals n cubed,

1^{3}=1

2^{3}=3+5

3^{3}=7+9+11

4^{3}=13+15+17+19

5^{3}=21+23+25+27+29

6^{3}=31+33+35+37+39+41

**From consecutive numbers to arithmetic series**

In arithmetic series, the difference between one term and the next term is constant. For example, the set of numbers -7, -3, 1, 5, 9, … forms an arithmetic series as each next number is obtained from the previous number by adding 4.

In general, terms of the arithmetic series can be written as the list of numbers

a, a+d, a+2d, a+3d, …

The concept of arithmetic series is broader than the concept of consecutive numbers, because the number which is added to or subtracted from each previous term (called the difference) of the series may not be a natural number.

As for consecutive numbers, for arithmetic series we can write a general rule that determines the n^{th} term of the arithmetic series,

n^{th} term equals to a+(n-1)d or more common notation a_{n}=a_{1}+(n-1)d, where

a_{n} is n^{th} term, a_{1} is the first term and d is the difference of the arithmetic series

Consider one more trick that can be applied to both a sequence of consecutive numbers and an arithmetic series. First, illustrate this trick with a partial numerical example.

**EXAMPLE:** What is the sum of first 100 even natural numbers?

**SOLUTION:** Consider first 100 consecutive even natural numbers,

2, 4, 6, 8, 10, …, 192, 194, 196, 198, 200.

Add pairs of two numbers as follows: the first number on the left and the first number on the right, the second number on the left and the second number on the right, … 50^{th} number on the left and 50^{th} number on the right – the sums in each case are 202 (see diagram).

There are exactly 50 such sums, so the sum of all consecutive even natural numbers from this list is

202×50=10100

In the same way you can calculate the sum of an arbitrary finite number of terms of the arithmetic series. Add the first number on the left and the first number on the right, the second number on the left and the second number on the right, … – the sums for each pair of numbers are 2a+(n-1)d (see diagram).

There are n2 such sums, therefore, the sum of n terms of the arithmetic series is

(2a+(n-1)d).$\frac{n}{2}$

**Quiz**

- The product of two consecutive integers is 2162. What are these integers?

**SOLUTION**: Let the smaller of two consecutive integers be n, then the greater integer is n+1. Therefore, their product equals to

n(n+1)=2162

Solve this equation:

n^{2}+n-2162=0

Using quadratic formula,

n_{1,2}=$\frac{-1±\sqrt{1^2-4⋅1⋅(-2162)}}{2⋅1}=\frac{-1±\sqrt{1+8648}}{2}=\frac{-1±93}{2}$=-47, 46

When n=-47, then n+1=-47+1=-46.

When n=46, then n+1=46+1=47.

**ANSWER:** -47 and -46 or 46 and 47

- Find four missing numbers in each sequence of consecutive numbers.

a) -56, -55, -54, …, -49, -48.

b) 17, 19, 21, …, 31.

c) 56, 51, …, 26, 21.

**SOLUTION**: a) The difference between any two consecutive numbers in this sequence is 1. The sequence is ascending, so the missing numbers are

-54+1=-53

-53+1=-52

-52+1=-51

-51+1=-50

b) The difference between any two consecutive numbers in this sequence is 2. The sequence is ascending, so the missing numbers are

21+2=23

23+2=25

25+2=27

27+2=29

c) The difference between any two consecutive numbers in this sequence is 5. The sequence is descending, so the missing numbers are

51-5=46

46-5=41

41-5=36

36-5=31

**ANSWER:** a) -53, -52, -51, -50 b) 23, 35, 27, 29 c) 46, 41, 36, 31

- How many natural numbers are from 83 to 118 inclusive? How many natural numbers are from 83 to 118 exclusive?

**SOLUTION**: We can represent number 118 as 83+35. There are 35+1=36 natural numbers from 83 to 118 inclusive and 35+1-2=34 natural numbers from 83 to 118 exclusive.

**ANSWER:** 36 natural numbers inclusive and 34 natural numbers exclusive

- What numbers are represented by a general rule 6k+11, where k is integer? Can you represent these numbers using another rule?

**SOLUTION:** Write down the first few members according to this rule:

- when k=3, 6k+11=6⋅3+11=29;
- when k=2, 6k+11=6⋅2+11=23;
- when k=1, 6k+11=6⋅1+11=17;
- when k=0, 6k+11=6⋅0+11=11;
- when k=-1, 6k+11=6⋅(-1)+11=5;
- when k=-2, 6k+11=6⋅-2+11=-1.

All these numbers leave the remainder of 5 when divided by 6. Therefore, the list of these consecutive numbers is

…,-7, -1, 5, 11, 17, 23, 29, …, 6k+5, …

By the way, recording all such consecutive numbers we managed to notice another rule that specifies numbers of this type: the numbers are of the form 6k+5, where k is integer.

**ANSWER**: …,-7, -1, 5, 11, 17, 23, 29, …, 6k+5, …

- How many consecutive odd numbers are between 872 and 976?

**SOLUTION:** The difference between two consecutive odd numbers is always 2.

The smallest possible odd number is number 873, the greatest possible odd number is number 975. We can represent number 975 as 873+102=873+2×51. This means we will add 2 to each previous number 51 times, so the number of consecutive numbers between 872 and 976 is 51+1=52.

**ANSWER:** 52

**Conclusions**

- Consecutive natural numbers differ from each other by 1. Adding 1 to a number, we get the successor of that number, subtracting 1 from a number we get predecessor of that number.
- Consecutive even numbers differ from each other by 2.
- Consecutive odd numbers differ from each other by 2.
- The general rule for all even numbers is 2k, where k is integer.
- The general rule for all odd numbers is 2k+1, where k is integer.
- Any consecutive numbers could be written using some general rule.
- The idea of consecutive numbers precedes the concept of arithmetic series.

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