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Last updated on October 16, 2025
In mathematics, a sequence is a set or list of numbers 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.
An arithmetic progression (AP) is a sequence in which each term is obtained by adding a constant to the previous term. 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 \(a_1\), and the nth term is denoted by \(a_n\).
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':
\( d = a_2 - a_1 = a_3 - a_2 = a_4 - a_3 = \dots = a_n - a_{n-1} \)
The nth term of an AP:
The nth term of an AP can be expressed as \( a_n = a + (n - 1)d \)
Where n is the number of terms
a is the first term
\(a_2\) is the second term
\(a_3\) is the third term
d is the common difference
\(a_n\) 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:
\( a_1, \; a_1 + d, \; a_1 + 2d, \; a_1 + 3d, \; \dots, \; a_1 + (n-1)d \), and the sum of these n terms is \(s_n\)
Then the sum of nth terms:
\( S_n = a_1 + (a_1 + d) + (a_1 + 2d) + (a_1 + 3d) + \dots + \big(a_1 + (n-1)d\big) \)
Now let's reverse the sequence and add corresponding terms. It can be written as:
\( S_n = a_n + (a_n - d) + (a_n - 2d) + (a_n - 3d) + \dots + \big(a_n - (n-1)d\big)\)
Adding these two equations:
\( S_n = a_1 + (a_1 + d) + (a_1 + 2d) + (a_1 + 3d) + \dots + \big(a_1 + (n-1)d\big) \)
\( S_n = a_n + (a_n - d) + (a_n - 2d) + (a_n - 3d) + \dots + \big(a_n - (n-1)d\big) \)
By adding these, we get: \( 2S_n = (a_1 + a_n) + (a_1 + a_n) + (a_1 + a_n) + \dots + (a_1 + a_n) \)
All the d terms are cancelled out here, so:
\( 2S_n = n(a_1 + a_n) \)
\( S_n = \frac{n(a_1 + a_n)}{2} \), let's consider it as equation 1
Substituting the formula to find an in equation 1, \( a_n = a_1 + (n - 1)d \)
Then, \( S_n = \frac{n}{2} \Big(a_1 + a_1 + (n - 1)d\Big) \)
\( S_n = \frac{n}{2} \Big[ 2a + (n - 1)d \Big] \), let’s consider this equation as 2
Equations 1 and 2 are used to find the sum of an arithmetic progression.
Hence, proved the arithmetic progression sum proof.
So, Sum of nth term: \( S_n = \frac{n}{2} \Big[ 2a + (n - 1)d \Big] \) and \( S_n = \frac{n}{2} (a + l) \)
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 \(a_n = a + (n - 1)d \).
For example, for the AP: 2, 4, 6, 8, 10,… we find the nth term.
Here, the first term a = 2 and the common difference d = 2.
\(a_n = a + (n - 1)d \)
\(= 2 + (n - 1)2\)
\(= 2 + 2n -2\)
\(a_n = 2n\)
To find the nth term of an AP, we use the formula \(a_n = a + (n - 1)d\).
For example, for the sequence 10, 20, 30, 40, …, here, \(a = 10\) and \(d = 10\).
So, the 5th term \(a_5 = 10 + (5 - 1)10 = 10 + 40 = 50\). Similarly, to find the 55th term, we can use the formula,
\(a_n = a + (n - 1)d \)
\(a_{55} = 10 + (55 - 1) \times 10 \)
\(= 10 + 54 × 10 \)
\(= 10 + 540 = 550. \)
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 an 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:
\(S_n = \frac{n}{2} \left( 2a + (n - 1)d \right) \)
When the nth term is known, the sum of n is calculated using: \(S_n = \frac{n}{2} (a + a_n) \)
Where 'n' is the number of terms,
'a' is the first term,
'd' is the common difference, and
\(a_n\) is the nth term.
Here are some tips and tricks for the children and their parents to master 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:
In our real world, we use arithmetic progression from basic counting to calculate the interest rate. Here are some real-life applications of arithmetic progression:
Exercise routines can be calculated using arithmetic progression. For example, increasing the number of push-ups by 2 each day: 2, 4, 6, 8… Distance or speed while cycling or running gradually increasing every week can follow an arithmetically progressive pattern.
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:
\(a_n = a + (n - 1)d \)
Here, \(a = 5\), \(d = 3\), and \(n = 10\)
So, \(a_{10} = 5 + (10 - 1) \times 3 \)
\(= 5 + 9 × 3\)
\(= 5 + 27 = 32\)
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 \(d = 7\)
\(a = 56\)
\(a_n = 196\)
So, \(n = \frac{a_n - a}{d} + 1 \)
\(= \frac{196 - 56}{7} + 1 \)
\(= 20 + 1 = 21 \)
Sum of first n terms: \(S_n = \frac{n}{2} (a + a_n) \)
\(= \frac{21}{2} (56 + 196) \)
\(= 21 × 126\)
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
\(a_n = a + (n - 1)d \)
So, \(a_{15} = 2 + (15 - 1) \times 15 \)
\(= 2 + 14 × 15 = 212\)
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 \(a_n = 205\)
\(a_n = a + (n - 1)d\)
\(205 = 7 + (n - 1) 6\)
\(205 - 7 = 6n -6\)
\(198 + 6 = 6n \)
\(204 = 6n\)
\(n = \frac{204}{6} = 34 \)
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.