**Definition of Arithmetic** Progression

An arithmetic progression is a numerical *sequence or series*, in which every consecutive value after the first is derived by adding a constant. This constant is called the *common** difference*. Such a numerical sequence is considered a progression because the last term in the sequence can be represented by an equation. So in order to find out the sum of n terms in Arithmetic Progression, let us first understand this specific kind of sequence, its terminologies, and formula.

## Explanation

During the course of our mathematic ventures, we often encounter numerical sequences such as these:

4, 6, 8, 10, 12 …

**and**

4, 7, 10, 13, 16, …

Studying the sequence of these numbers, we can ascertain a pattern and the next expected value in the numerical sequence. In the first sequence, each of the succeeding terms is derived by adding two; in the latter, three is added each time, after the initial term. Importantly, it is not necessary that the sequences only be derived by addition to constituting an arithmetic progression. A constant that is subtracted from each term after the initial term to derive the consecutive number may also be considered an academic progression.

## The Sum of N Terms in Arithmetic Progression

Prior to deriving a formula to calculate the n^{th }term in arithmetic progression, let us consider how the sum of all natural numbers between 1-100 can be derived without a formula.

In this sequence, the sum of numbers can be represented as such:

*Sum*= 1+2+3+4+5+6….+97+98+99+100

Even when the order is reversed, the sum does not change:

*Sum*= 100+99+98+97….6+5+4+3+2+1

Hence, when both equations 1 and 2 are added, we get:

2 x *Sum = *(100+1) + (99+2) + (98+3) + (97+4) + … (4+97) + (3+98) + (2+99) + (1+100)

2 x *Sum *= 101+ 101 + 101 + 101 + … (4+97) + (3+98) + (2+99) + (1+100)

2 x *Sum *= 101 (100 terms)

2*Sum* = 101(100)

*Sum *=

*Sum *= 5,050

This method can be used to calculate the sum of natural numbers like 1000, 10,000 or even 100,000. Hence, knowing the last term in the sequence, this method can be used to derive the formula needed to figure out the n^{th }term in any given sequence.

If we express the first term in the academic progression as *a, *the common difference between each consecutive term as *d,* and the last term as *l.*

*Sum = a + (a + d) + (a + 2d) + (a + 3d)…+ (l – 3d) + (l – 2d) + (l – d) + l*

Where *l *= **a + (n – 1) d**

In reverse order, the sum remains the same:

*Sum*=*l + (l – d) + (l – 2d) + (l – 3d) + … (a + 3d) + (a + 2d) + (a + d) + a*

Adding equations a and b, we get:

2 x *Sum = (a + l) + [(a +d) + (l – d)] … + [(l – d) + (a + d)] + (l + a)]*

- 2 x
*sum*=*(a + l) + (a + l)…+ (a + l) + (a + l)* - 2 x
*sum*= n x (a + l) *sum =*

Substituting *l *with the previous equation above (where *l = a + (n – 1)d): *

*Sum of n terms of AP =*

**Arithmetic Progression**

In an arithmetic progression, not all numbers in the sequence may be known or given. To find the n^{th} term, the equation: ** a + (n – 1)d **is used.

**Examples**

Let’s take the first sequence as an example: *4, 6, 8, 10, 12,…*

The table below is used to visualize the components needed to fulfill the arithmetic progression:

Terms |
n |
(n-1) |
d |

4 | 1 | 0 | – |

6 | 2 | 1 | 2 |

8 | 3 | 2 | 2 |

10 | 4 | 3 | 2 |

12 | 5 | 4 | 2 |

In the first column, we list all given numbers from the sequence in their order of appearance. Column 2, *n*, denotes the place where the terms appear in the sequence. Column 3, *d,* contains the common difference between each term, which in this case is the addition of 2 to each term, to derive the next consecutive term. Below we input the information from the tale into the equation: *a + (n-1)d**.*

For the second sequence, the arithmetic progression may be expressed as:

By now, you must be wracking your brain for what possible use you might have of such a mathematical concept. Interestingly, arithmetic progression is so ubiquitous that we are constantly surrounded by it in our daily lives. On the rare but lucky occasion that your local bus is running on time, you can thank arithmetic progression. If we take the first arrival of the bus at the stop as the initial term and use every consecutive arrival at a constant interval of time to derive the* common difference*. Each return to the same stop is considered a value of *n*. Knowing the initial time, the common difference, and *n*, we can predict the time at which the next bus or the bus after a certain interval will arrive!

### KEY TAKEAWAYS

An arithmetic progression** **is the sequencing of numbers in which the consecutive number is derived through a sum, and in which there is a *common difference* between two consecutive terms. The n^{th} term can be derived using the formula ** a + (n-1)d, **where

*a*is the initial term,

*n*is the numerical order in which the n

^{th }term appears, and

*d*is the common difference between two consecutive terms.

In an arithmetic progression, it is possible to figure out the ** sum of n terms** manually, using

*Sum of n terms in Arithmetic Progression =*

Also check out our article on Geometry formulas here.