Last updated on July 4th, 2025
In mathematics, a sequence is a set or list of numbers that are arranged in a particular order. An arithmetic progression (AP) is a sequence where the terms are arranged in such a way that the difference between any two successive terms is constant. In this article, we will learn more about arithmetic progression with examples.
A sequence where a term is found by adding a constant to its previous term is called an arithmetic progression. In other words, the difference between two consecutive terms in an AP will be the same. The difference is known as the common difference (d), and the first term is denoted by a or a1, and the nth term is denoted by an. For example, 3, 6, 9, 12, 15, … is an arithmetic progression. Here, a is 3, and the constant d is also 3. In its general form, AP can be expressed as a, a + d, a + 2d,…, a + (n - 1)d.
There are mainly two types of progressions: arithmetic and geometric (GP). Here, we will be discussing the difference between AP and GP:
Arithmetic Progression | Geometric Progression |
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Example: 5, 10, 15, 20, 25, … | Example: 3, 6, 12, 24, …. |
Key formulas for arithmetic progression for calculating common difference, the nth term, and the sum of terms.
The common difference in AP:
The common difference of a sequence is the difference between two consecutive terms, and it is denoted by d:
d = a2 - a = a3 - a2 = a4 - a3 = … = an - an-1
The nth term of an AP:
The nth term of an AP can be expressed as an = a + (n - 1)d
Where n is the number of terms
a is the first term
a2 is the second term
a3 is the third term
d is the common difference
an is the nth term of the sequence
Now we will learn how to find the sum of the first n terms of an AP. So let’s consider the first n terms of an AP as a1, a1 + d, a1 + 2d, a1 + 3d, ….., a1 + (n - 1)d, and the sum of these n terms is sn
Then the sum of nth terms:
Sn = a1 + (a1 + d) + (a1 + 2d) + (a1 + 3d) + ….., + (a1 + (n - 1)d)
Now let's reverse the sequence and add corresponding terms. It can be written as:
Sn = an + (an - d) + (an - 2d) + (an - 3d) + ….., + (an + (n - 1)d)
Adding these two equations:
Sn = a1 + (a1 + d) + (a1 + 2d) + (a1 + 3d) + ….., + (a1 + (n - 1)d)
Sn = an + (an - d) + (an - 2d) + (an - 3d) + ….., + (an + (n - 1)d)
By adding these, we get: 2sn = (a1 + an) + (a1 + an) + (a1 + an) + …. + (a1 + an)
All the d terms are cancelled out here, so:
2sn = n(a1 + an)
Sn = n(a1 + an)/2, let's consider it as equation 1
Substituting the formula to find an in equation 1, an = a1 + (n - 1)d
Then, Sn = n/2 (a1 + a1 + (n - 1)d)
= n/2 [(2a + (n - 1)d], let’s consider this equation as 2
Equations 1 and 2 are used to find the sum of an arithmetic progression.
So, Sum of nth term: sn = n/2 (2a + (n - 1)d) and sn = n/2(a + l)
The nth term of an AP is used to find any term in the sequence without listing all the previous terms. There is a formula that we can use to find the nth term, which is an = a + (n - 1)d.
For example, for the AP: 2, 4, 6, 8, 10,… we find the nth term. Here, a = 2, d = 2
an = a + (n - 1)d
= 2 + (n - 1)2
= 2 + 2n -2
an = 2n
To find the nth term of an AP, we use the formula an = a + (n - 1)d. For example, for the sequence 10, 20, 30, 40, …, here, a = 10 and d = 10.
So, the 5th term a5 = 10 + (5 - 1)10 = 10 + 40 = 50. Similarly, to find the 55th term, we can use the formula,
an = a + (n - 1)d
a55 = 10 + (55 - 1) 10
= 10 + 54 × 10
= 10 + 540 = 550.
So, the 55th term of this AP is 550.
The arithmetic sequence explicit formula is the nth term of an AP, it is used to find any term of the sequence.
Now we will learn to find the sum of arithmetic progression. The sum of the first ‘n’ terms can be calculated with the help of the formula mentioned below.
When the nth term is unknown, the sum of n is calculated using the formula:
Sn = n/2 (2a + (n - 1)d)
When the nth term is known, the sum of n is calculated using: Sn = n/2 (a + an)
Where,
n is the number of terms
a is the first term
d is the common difference
a is the first term
an is the nth term
In our real world, we use AP from basic counting to calculate the interest rate. Here, we will learn some applications of arithmetic progression.
When working on arithmetic progression, students often make mistakes, which leads to errors. Here are some common mistakes and the ways to avoid them:
Find the 10th term of AP, where the first term is 5 and the common difference is 3
The 10th term of the AP is 32
The nth term of an AP is calculated by:
an = a + (n - 1)d
Here, a = 5, d = 3, and n = 10
So, a10 = 5 + (10 - 1) 3
= 5 + 9 × 3
= 5 + 27 = 32
Find the common differences between AP 2, 5, 8, 11,…
The common difference is 3
The common difference is the difference between the two consecutive terms.
d = 5 - 2 = 3
Find the sum of all multiples of 7 between 50 and 200?
The sum of all the multiples of 7 between 50 and 200 is 2646
The first multiple of 7 that is greater than or equal to 50 is 56
The last multiple of 7 that is less than or equal to 200 is 196
Here, the d = 7
a = 56
an = 196
So, n = an- a/d + 1
= (196 - 56 / 7) + 1
= 20 + 1 = 21
Sum of first n terms: Sn = n/2 (a + an)
= 21/2 (56 + 196)
= 21 × 126
2646
Find the 15th term of the AP where a = 2 and d = 15?
The 15th term is 212
Here, the first term (a) is 2
The common difference (d) is 15
an = a + (n - 1)d
So, a15 = 2 + (15 - 1)15
= 2 + 14 × 15 = 212
Find the number of terms in the AP 7, 13, 19, …, 205.
The number of terms in the AP 7, 13, 19, … , 205 is 34.
Here a = 7, d = 6, and an = 205
an = a + (n - 1)d
205 = 7 + (n - 1) 6
205 - 7 = 6n -6
198 + 6 = 6n
204 = 6n
n = 204 / 6 = 34
So, the number of terms is 34.
Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.
: He loves to play the quiz with kids through algebra to make kids love it.