Arithmetic Sequence (original) (raw)
Last Updated : 23 Apr, 2026
An arithmetic sequence or progression is defined as a sequence of numbers in which the difference between one term and the next term remains constant.
**For example, the given sequence below has a common difference of 1.
1 2 3 4 5 . . . n
⇑ ⇑ ⇑ ⇑ ⇑ . . .
1st 2nd 3rd 4th 5th . . . nth Terms
The Arithmetic Sequence can be represented as
Arithmetic Sequence / Arithmetic Progression
The above image has a common difference of 2.
**Note: A negative common difference causes the sequence to decrease, and this is entirely valid in an arithmetic progression (AP).
General Formula for the Nth Term
The general formula to find the nth term of an arithmetic sequence is:
an = a + (n − 1) ⋅ d
Where:
- an = nth term,
- a = first term,
- d = common difference,
- n = term number.
**Example: sequence below
Let's consider an example of Arithmetic Sequence 6, 16, 26, 36, 46, 56, 66, . . .
| Term (n) | Expression | Value |
|---|---|---|
| a1 | 6 + (1 - 1) ⋅ 10 | 6 |
| a2 | 6 + (2 -1) ⋅ 10 = 6 + 10 | 16 |
| a3 | 6 + (3 − 1) ⋅ 10 = 6 + 10 + 10 | 26 |
| a4 | 6 + (4 - 1) ⋅ 10 = 6 + 10 + 10 + 10 | 36 |
| a5 | 6 + (5 − 1) ⋅ 10 = 6 + 10 + 10 + 10 + 10 | 46 |
| a6 | 6 + (6 - 1) ⋅ 10 = 6 + 10 + 10 + 10 + 10 + 10 | 56 |
| a7 | 6 + (7 − 1) ⋅ 10 = 6 + 10 + 10 + 10 + 10 + 10 + 10 | 66 |
**General Formula for the above sequence: Substituting the values a = 6 and d = 10 into the formula:
an = 6 + (n - 1) × 10
This formula can be used to find any term of the sequence directly. For example:
- a10 = 6 + (10 − 1) ⋅ 10 = 6 + 90 = 96,
- a22 = 6 + (22 − 1) ⋅ 10 = 6 + 210 = 216.
Sum of terms in Arithmetic Sequence
The sum of terms of an Arithmetic Sequence is called an **Arithmetic Series.
Suppose the first term of the arithmetic series is **a and the common difference is **d then the sum of the **n term of this arithmetic series is given using the formula,
Sn= n/2 [2a + (n - 1)d]
If the common difference of the arithmetic series is not given but the last term of the series is given (say l). Then its sum is calculated as,
Sn= n/2 [a + l]
**Example: A tree fruits five apples in the first year and at each successive year it has 2 more apples than the last year find the total apple the tree bears at the end of six years.
**Solution:
Apple bear by tree in first year (a) = 5
Yearly Increase in apple bear by the tree (d) = 2
Time Period = 6 years
Total apple at the end of six years in the tree.
Sn = n/2 [2a + (n - 1)d]
⇒ Sn = 6/2(2(5) + (6 - 1)(2))
⇒ Sn = 6/2 (10 + 10)
⇒ Sn = 3(20)
⇒ Sn = 60Thus, the total number of apples the tree bears at the end of 6 years is 60 apples.
Recursive Formula for Arithmetic Sequence
The nth term of an arithmetic sequence can also be defined recursively. Each term is obtained by adding a common difference _d to the previous term.
As we know, nth term is: an = a + (n - 1) × d
thus, (n - 1)th term can be given by an-1 = a + (n - 2) × d
Now,
an = a + (n - 1) × d= a + (n − 2)d + d [We rewrite (n - 1) as: (n - 1) = (n - 2) + 1**]**
= an-1 + d
**Recursive Formula: an = an−1 + d
Important Formulas & Notations in Arithmetic Progression
The notations used in arithmetic progression are :
- First Term ⇢ **a
- Common Difference ⇢ d
- Nth term ⇢ **a n
- Sum of First n Terms ⇢ **S n
The following table contains formulas related to Arithmetic Progression.
| Common Difference | __a_n+1 __- a_n _= d |
|---|---|
| General Form of AP | a, a + d, a + 2d, a + 3d, . . . |
| nth term of AP | an = a + (n – 1) × d |
| Sum of n terms in AP | S = n/2[2a + (n − 1) × d] |
| Sum of all terms in a finite AP with the last term as ‘l’ | n/2(a + l) |
Related Articles:
Examples of Arithmetic Sequence
**1. Sequence of Natural Numbers
1, 2, 3, 4, 5, 6, . . .
- First term a1 = 1
- Common difference (d): 2 - 1 = 1.
**2. Sequence of Even Numbers
2, 4, 6, 8, 10, 12, . . .
- First term a1 = 2
- Common difference (d): 4 - 2 = 2.
**3. Sequence of Odd Numbers
1, 3, 5, 7, 9, 11, . . .
- First term a1 = 1
- Common difference (d): 3 - 1 = 2.
**4. Negative Arithmetic Sequence:
−5, −10, −15, −20, -25, -30, . . .
- First term a1 = -5
- Common difference (d): -10 - (-5) = -10 + 5 = -5.
Solved Problems
**Question 1: Find the AP if the first term is 15 and the common difference is 4.
**Solution:
As we know,
a, a + d, a + 2d, a + 3d, a + 4d, …Here, a = 15 and d = 4
= 15, (15 + 4), (15 + 2 × 4), (15 + 3 × 4), (15 + 4 × 4),
= 15, 19, (15 + 8), (15 + 12), (15 + 16), …
= 15, 19, 23, 27, 31, …and so on.So the AP is 15, 19, 23, 27, 31...
**Question 2: Find the 20th term for the given AP: 3, 5, 7, 9, …
**Solution:
Given, 3, 5, 7, 9, 11, . . .
a = 3, d = 5 – 3 = 2, n = 20Thus, an = a + (n − 1)d
⇒ a20 = 3 + (20 − 1)2
⇒ a20 = 3 + 38
⇒ a20 = 41Here 20th term is a20 = 41
**Question 3: Find the sum of the first 20 multiples of 5.
**Solution:
First 20 multiples of 5 are 5, 10, 15, . . . , 100.
Here, it is clear that the sequence formed is an AP where,
a = 5, d = 5, an = 100, n = 20.
Thus, Sn = n/2 [2a + (n − 1) d]
⇒ Sn = 20/2 [2 × 5 + (20 − 1)5]
⇒ Sn = 10 [10 + 95]
⇒ Sn = 1050
Unsolved Problems
**Question 1: Find the 12th term of the arithmetic sequence: 7, 11, 15, 19, …
**Question **2: Determine the sum of the first 15 terms of the sequence: 5, 10, 15, 20, …
**Question **3: The 5th term of an arithmetic sequence is 18, and the 12th term is 39. Find the first term and the common difference.
**Question **4: How many terms are there in the sequence 8, 14, 20, …, 92?
**Question **5: The sum of the first n terms of an AP is 3n² + 5n. Find the first term, common difference, and the 10th term.