A sequence of
Integers where each integer is one more than the previous integer in the
sequence are called Consecutive Integers. The consecutive integers can
be represented by n, n+1, n+2,....

### Consecutive Even Integers:

The
sequence of integers which start with an even number and each integer
in the sequence is 2 more than the previous integer in the sequence are
called Even Consecutive Integers. Example 2, 4, 6.......

### Consecutive Odd Integers:

The
sequence of integers which start with an odd integer and each of the
following integers in the sequence are two more than the previous integer
in the sequence are called as Consecutive Odd Integers. Example 35,
37, 39, .....

We must be able to represent a
sequence of consecutive integers. If the first number in the sequence is
n, then depending upon what sort of consecutive integers we need, we
decide on what constant to add to it. For example,

n, n+1, n+2, .... Represents a sequence of consecutive integers

n, n+2, n+4,..... Represents a sequence of odd integers if n is odd and a sequence of even integers if n is even.

n, n+5, n+10,..... represents a sequence of consecutive multiples of 5 if the first integer is a multiple of 5.

We
must also be able to reconstruct the required consecutive integer
sequence, once we know the value of n. For example when n = 6

If we have to find three consecutive integers then they are 6, 7, 8

If we have to find four consecutive even numbers, 6, 8, 10, 12

If we have to find three consecutive multiples of three then 6, 9, 12

### Examples on Consecutive Integers:

Given below are some examples on consecutive integers.

**Example 1:** The sum of the greatest and least of a set of three consecutive integers is 120. What are the three integers in the sequence?

**Solution:**

Let the three consecutive integers in the sequence be x, x+1, x+2

The sum of the consecutive integers is 120.

x + x +1 + x+2 = 120

3x + 3 = 120

Solve for x

3x = 120 - 3

x = $\frac{117}{3}$ = 39

x+1 = 40

x+2 = 41

Hence, the three consecutive integers are 39 ,40, 41

**Example 2:**

If the difference between the cubes of two consecutive integers is 217, then find the integers?

**Solution: **

Let the two consecutive integers be n and n+1. Then the difference between their cubes is

n³ - (n+1)³ = 217

Now we have to solve for n.

n³ - [n³ +1 + 3n(n+1)] = 217 Since (n+1)³ = n³ + 1 + 3n(n+1)

n³ - n³ - 1 - 3n² - 3n = 217 Eliminating the braces

3n² + 3n - 216 = 0 Simplify

n² + n -72 = 0 multiplying by $\frac{1}{3}$

(n + 9)(n - 8) = 0 Factorizing

n = -9 or n = 8

Now we have to decide which one to take up. Let us try

n = 8 then the next integer is 9

the difference between the cubes of 8 and 9 is 9³ - 8³ = 217

Let n = -9. The next consecutive integer is n+1 = -8.

(-9)³ - (-8)³ = -217 which is not what is given.

Hence the two consecutive integers are 9, 8