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 below sequence 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 ProgresionThe 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).
The general formula to find the nth term of an arithmetic sequence is:
an = a1 + (n − 1) ⋅ d
Where:
- an = nth term,
- a = first term,
- d = common difference,
- n = term number.
Example: Nth Term of Arithmetic Sequence
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.
Examples of Arithmetic Sequence
Here are some examples of arithmetic sequences,
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.
Read more: How to Find the Nth term of Arithmetic Sequence?
Sum of terms in Arithmetic Sequence
The sum of terms of an Arithmetic Sequence is calledan 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 five 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 = 60
Thus, the apple in the tree at the end of six-year is 60 apples
The nth term of an Arithmetic Sequence can be defined recursively as the next term can always be obtained by adding a common difference to the preceding term, the following derivation can be used to illustrate the same thing.
As we know, nth term of the Arithmetic Sequence is given by
an = a + (n - 1) × d
thus, (n - 1)th term can be given by
an-1 = a + (n - 1 - 1) × d
an-1 = a + (n - 2) × d
Thus,
an = a + (n - 1) × d
⇒ an = a + (n - 1 - 1 + 1) × d
⇒ an = a + (n - 2 + 1) × d
⇒ an = a + (n - 2) × d + d
⇒ an = an-1 + d
The notations used in arithmetic progression are :
The following table contains formulas related to Arithmetic Progression.
Common Difference | an+1 - an = 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) |
|---|
Solved Problems on Arithmetic Sequence
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 = 20
Thus, an = a + (n − 1)d
⇒ a20 = 3 + (20 − 1)2
⇒ a20 = 3 + 38
⇒ a20 = 41
Here 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
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