Introduction
Finding a formula for the nth term for a specific kind of sequence called a progression is possible. The most popular mathematical progression, with simple formulas, is the arithmetic progression (AP). A good example of an arithmetic progression (AP) is the sequence 3, 8, 13, 18, 23, 28, 33…, which follows a pattern in which each number is obtained by adding 5 to the previous term.
Let us look at the concept of an arithmetic progression, its formula, the sum of the first n term, and several examples with solutions in this article.
What is Arithmetic Progression?
Definition
An “arithmetic progression” (AP) is a series of numbers when any two subsequent numbers have a constant difference. Arithmetic Sequence is another name for it. An arithmetic sequence is a succession of numbers in which, for each pair of consecutive terms, the second number is produced by adding the first one by a fixed number.
For example, in the sequence 4, 7, 10, 13, 16, 19, 22, 25, 28, and 31, the first term is 4, while the common difference between terms is 3.
Let’s look at the table below to show the general form of an arithmetic sequence. Suppose that a is the first term, n is the position of the term, and d is a common difference.
| Position of terms | Nth term | Value of Nth Term | |
| 1 | a1 | a | a + ( 1 – 1 ) d |
| 2 | a2 | a + d | a + ( 2 – 1 ) d |
| 3 | a3 | a + 2d | a + ( 3 – 1 ) d |
| 4 | a4 | a + 3d | a + ( 4 – 1 ) d |
| 5 | a5 | a + 4d | a + ( 5 – 1 ) d |
| 6 | a6 | a + 5d | a + ( 6 – 1 ) d |
| 7 | a7 | a + 6d | a + ( 7 – 1 ) d |
| . | . | . | . |
| . | . | . | . |
| . | . | . | . |
| n | an | an = a + ( n – 1 ) d |
The Formula of Arithmetic Progression
In algebra, arithmetic progression is the arrangement of numbers such that there is a constant difference between terms. An arithmetic sequence is a succession of numbers in which, for each pair of consecutive terms, the second number is produced by adding the first one by a fixed number.
The formula to find the nth term of an arithmetic progression is given by,
an= a + ( n – 1 ) d
where an = nth term,
a = first term,
n = position of the term
d = common difference
Nth Term of Arithmetic Progression
The formula an = a + ( n – 1 ) d is used to get the general term (or) nth term of an arithmetic progression (AP) whose first term is a, and the common difference is d.
For example, we have the sequence 5, 8, 11, 14, 17, 20, 23, and 26.
Finding the general term or the nth term, we will substitute a =5 as the first term and d = 3 as the common difference in the formula. So, we have,
an = a + ( n – 1 ) d
an = 5 + ( n – 1 ) 3
an = 5 + 3n – 3
an = 3n + 2
Thus, the general term of the sequence 5, 8, 11, 14, 17, 20, 23, 26,… is an = 3n + 2.
Obtaining the nth term of the arithmetic progression is helpful to identify the value of the other terms in the sequence, let us say the 30th term, 45th term, 90th term, 100th term, etc., without manually listing the sequence.
For example, let us use the computed general term an = 3n + 2 and identify the 30th term, 45th term, 90th term, and 100th term.
| when n = 30 an = 3n + 2 a30 = 3(30) + 2 a30 = 90 + 2 a30 = 92 | when n = 45 an = 3n + 2 a45 = 3(45) + 2 a45 = 135 + 2 a45 = 137 | when n = 90 an = 3n + 2 a90 = 3(90) + 2 a90 = 270 + 2 a90 = 272 | when n = 100 an = 3n + 2 a100 = 3(100) + 2 a100 = 300 + 2 a100 = 302 |
Hence, the 30th term is 92, the 45th term is 137, the 90th term is 272, and the 100th term is 302 of the sequence 5, 8, 11, 14, 17, 20, 23, 26,…
Sum of the First n Terms of Arithmetic Progression
The formula for the first n terms of an arithmetic progression is
Formula (First n Numbers in an AP): Sn = n/2 [ 2a + ( n – 1 ) d ]
where Sn = sum of the n terms
n = total terms
a = first term
d = common difference
Let us consider adding the first 30 numbers in the sequence, 2, 6, 10, 14, 18, …
In the example given, the first term is 2 or a = 2; the common difference is 4 or d = 4, and n = 20. Substituting the given into the formula we have,
S30 = 30/2 [ 2 (2) + (30 – 1) ( 4 ) ]
S30 = 15 [ 4 + ( 29 )( 4 ) ]
S30 = (15) ( 4 + 116)
S30 = ( 15 ) ( 120 )
S30 = 1800
Therefore, the sum of the first 30 numbers in the sequence 2, 6, 10, 14, 18, … is 1800.
For another example, let us say we want to get the sum of the first 50 natural numbers.
The arithmetic progression for the scenario is 1, 2, 3, 4, 5, … 45, 46, 47, 48, 39 50.
The first term is 1 or a = 1, the common difference is 1 or d = 1, and n = 50.
Applying the values to the existing formula,
S50 = 50/2 [ 2 (1) + ( 50 – 1 ) ( 1 ) ]
S50 = 25 [ 2 + ( 49 )(1 ) ]
S50 = (25) ( 2 + 49 )
S50 = ( 25 ) ( 51 )
S50 = 1275
When the first and last terms are given, the formula of the sum of the first n terms of the arithmetic progression is given by
Sn = n/2 ( first term + last term )
For example, let us use the previously given sum of the first 50 natural numbers. Since the given tells that the first term is 1 and the last is 50, the solution will be,
Sn = n/2 ( first term + last term )
S50 = 50/2 ( 1 + 50 )
S50 = 25 ( 51 )
S50 = 1275
Thus, the sum of the first 50 natural numbers is 1275.
More Examples
Example 1
Find a25 in each of the following arithmetic progression.
( a ) 10, 15, 20, 25, 30, 35 …
( b ) 12, 20, 28, 36, 44, 52 …
Solution
( a ) 10, 15, 20, 25, 30, 35 …
The common difference is 5 or d = 5, and the first term is 10 or a = 10.
an = a + ( n – 1 ) d
a25 = 10 + ( 25 – 1 ) 5
a25 = 10 + ( 24 ) 5
a25 = 10 + 120
a25 = 130
( b ) 12, 20, 28, 36, 44, 52 …
The common difference is 8 or d = 8, and the first term is 12 or a = 12.
an = a + ( n – 1 ) d
a25 = 12 + ( 25 – 1 ) 8
a25 = 12 + ( 24 ) 8
a25 = 12 + 192
a25 = 204
Example 2
Find the value of n.
( a ) d = 5, a = 4, and an = 99
( b ) d = 9, a = 2, and an = 308
( c ) d = – 2 , a = 25, and an = -9
Solution
( a ) d = 5, a = 4, and an = 99
Let us substitute the given into the formula.
a + ( n – 1 ) d = an
4 + ( n – 1 ) 5 = 99
4 + 5n – 5 = 99
5n – 1 = 99
5n = 99 + 1
5n = 100
n = 100 / 5
n = 20
( b ) d = 9, a = 2, and an = 308
Let us substitute the given into the formula.
a + ( n – 1 ) d = an
2 + ( n – 1 ) 9 = 308
2 + 9n – 9 = 308
9n – 7 = 308
9n = 308 + 7
9n = 315
n = 315 / 9
n = 35
( c ) d = – 2 , a = 25, and an = -9
Let us substitute the given into the formula.
a + ( n – 1 ) d = an
25 + ( n – 1 ) (-2) = -9
25 – 2n + 2 = -9
-2n + 27 = -9
-2n = -9 – 27
-2n = -36
n = -36 / -2
n = 18
Example 3
Find the general term or the explicit formula of the given arithmetic progression.
( a ) -4, 1, 6, 11, 16, 21, 26, 31, 36, 41, 46, 51, 56, 61, …
( b ) -19, -7, 5, 17, 29, 41, 53, 65, 77, 89, 101, 113, 125, …
( c ) 10, 4, -2, -8, -14, -20, -26, -32, 38, -44, -50,…
Solution
( a ) -4, 1, 6, 11, 16, 21, 26, 31, 36, 41, 46, 51, 56, 61, …
Let us use a = -4 and d = 5 into the formula.
an = a + ( n – 1 ) d
an = -4 + ( n – 1 ) 5
an = -4 + 5n – 5
an = 5n – 9
( b ) -19, -7, 5, 17, 29, 41, 53, 65, 77, 89, 101, 113, 125, …
Let us use a = -19 and d = 12 into the formula.
an = a + ( n – 1 ) d
an = -19 + ( n – 1 ) 12
an = -19 + 12n – 12
an = 12n – 31
( c ) 10, 4, -2, -8, -14, -20, -26, -32, 38, -44, -50,…
Let us use a = 10 and d = – 6 into the formula.
an = a + ( n – 1 ) d
an = 10 + ( n – 1 ) (-6)
an = 10 – 6n + 6
an = -6n + 16
Example 4
Find the sum.
( a ) S30 of the AP 15, 20, 25, 30, 35, 40,…
( b ) S25 of the AP 8, 15, 22, 29, 36, 43,…
( c ) S14 of the AP 20, 17, 14, 11, 8, 5, 2,…
Solution
( a ) S30 of the AP 15, 20, 25, 30, 35, 40,…
Let us plug in a = 15, n = 30, and d =5 into the formula.
S30 = 30/2 [ 2 (15) + ( 30 – 1 ) ( 5 ) ]
S30 = 15 [ 30 + ( 29 )(5 ) ]
S30 = (15) ( 30 + 145 )
S30 = ( 15 ) ( 175 )
S30 = 2625
( b ) S25 of the AP 8, 15, 22, 29, 36, 43,…
Let us plug in a = 8, n = 25, and d =7 into the formula.
S25 = 25 /2 [ 2 ( 8 ) + ( 25 – 1 ) ( 7 ) ]
S25 = 12.5 [ 16 + ( 24 )(7 ) ]
S25 = ( 12.5 ) ( 16 + 168 )
S25 = ( 12.5) ( 184 )
S25 = 2300
( c ) S14 of the AP 20, 17, 14, 11, 8, 5, 2,…
Let us plug in a = 17, n = 14, and d =-3 into the formula.
S14 = 14 /2 [ 2 ( 20 ) + ( 14 – 1 ) ( -3 ) ]
S14 = 7 [ 40 + ( 13 )(-3 ) ]
S14 = ( 7 ) ( 40 -39 )
S14 = ( 7) ( 1 )
S14 = 7
Example 5
In the AP 5, 12, 19, 26, 33, 40, 47, 54, 61, 68, 75, …, find a45 and S45.
Solution
Let’s input the given values, a = 5, n = 45, and d = 7, into the equation.
an = a + ( n – 1 ) d
a45 = 5 + ( 45 – 1 ) 7
a45 = 5 + (44) 7
a45 = 5+ 308
a45 = 313
| S45 = 45 /2 [ 2 ( 5 ) + ( 45 – 1 ) ( 7 ) ] S45 = 22.5 [ 10 + ( 44 )(7 ) ] S45 = ( 22.5 ) ( 10 + 308 ) S45 = ( 22.5) ( 318 ) S45 = 7155 | Alternative solution Sn = n/2 ( first term + last term ) S45 = 45/2 ( 5 + 313 ) S50 = 22.5 ( 318 ) S50 = 7155 |
Therefore the 45th term in the sequence is 313, and the sum of the first 45 terms is 7155.
Summary
An “arithmetic progression” (AP) is a series of numbers when any two subsequent numbers have a constant difference. Arithmetic Sequence is another name for it. An arithmetic sequence is a succession of numbers in which, for each pair of consecutive terms, the second number is produced by adding the first one by a fixed number.
The formula to find the nth term of an arithmetic progression is given by,
an= a + ( n – 1 ) d
where an = nth term,
a = first term,
n = position of the term
d = common difference
Given by is the formula for the first n terms of an arithmetic progression.
Formula (First n Numbers in an AP): Sn = n/2 [ 2a + ( n – 1 ) d ]
where Sn = sum of the n terms
n = total terms
a = first term
d = common difference
When the first and last terms are given, the formula of the sum of the first n terms of the arithmetic progression is given by
Sn = n/2 ( first term + last term )
Frequently Asked Questions on Arithmetic Progression (AP)
What is the difference between finite arithmetic sequence and infinite arithmetic sequence?
If an arithmetic sequence contains a finite or limited number of terms, it is finite; otherwise, it is infinite.
For example, { 10, 13, 16, 19, 22, 25, 28, 31 } is a finite arithmetic sequence since it has a limited number of terms and the last term.
An example of an infinite arithmetic sequence is { 2, 6, 10, 14, 18, 22, 26, 30, … }
What is the formula to calculate arithmetic progression?
An “arithmetic progression” (AP) is a series of numbers when any two subsequent numbers have a constant difference.
An arithmetic progression’s nth term can be determined using the following formula:
an= a + ( n – 1 ) d
where an = nth term,
a = first term,
n = position of the term
d = common difference
What distinguishes geometric progression from arithmetic progression?
In algebra, an arithmetic progression (AP) is the arrangement of numbers such that there is a constant difference between terms. An arithmetic sequence is a succession of numbers in which, for each pair of consecutive terms, the second number is produced by adding the first one by a fixed number.
The formula to find the nth term of an arithmetic progression is given by,
an= a + ( n – 1 ) d
where an = nth term,
a = first term,
n = position of the term
d = common difference
For example, 3, 7, 11, 15, 19, 23,… is an arithmetic progression with a common difference of 4.
7 – 3 = 4
11 – 7 =4
15 – 11 =4
19 – 15 = 4
A type of sequence known as geometric progression (GP) is one in which each subsequent term is obtained by multiplying each previous term by a fixed number, or “common ratio.” Another name for this progression is a geometric sequence of numbers.
The following formula can be used to find the nth term in a geometric progression:
an= a (r)n-1
where an = nth term,
a = first term,
n = position of the term
r = common ratio
For example, 2, 6, 18, 54, 162, 486,… is a geometric progression with a common ratio of 3.
6 / 2 = 3
18 / 6 = 3
54 / 18 = 3
162 / 54 =3
What is the graph of an arithmetic progression?
A straight line is a graph of an arithmetic progression with its slope as the common difference.
How to calculate the sum of the first n numbers in an arithmetic progression?
The sum of the first n term of an arithmetic progression is given by
Formula (First n Numbers in an AP): Sn = n/2 [ 2a + ( n – 1 ) d ]
where Sn = sum of the n terms
n = total terms
a = first term
d = common difference
Let us consider adding the first 15 numbers in the sequence, 3, 12, 21, 30, 39, …
In the example given, the first term is 3 or a = 3, the common difference is 9 or d = 9, and n = 15. Substituting the given into the formula we have,
S15 = 15/2 [ 2 (3) + ( 15 – 1 ) ( 9 ) ]
S15 = 7.5 [ 6 + ( 14 )( 9 ) ]
S15 = (7.5) ( 6 + 126 )
S15 = ( 7.5 ) ( 132 )
S15 = 990
Therefore, the sum of the first 15 numbers in the sequence 3, 12, 21, 30, 39, …
is 990.
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