Arithmetic Progressions Class 10th Notes - Free NCERT Class 10 Maths Chapter 5 Notes - Download PDF

An arithmetic progression is a sequence of numbers where the numbers follow an order and have a constant difference between them. An arithmetic progression is also known as an arithmetic sequence, where the difference between the previous and next term is the same. In an arithmetic progression, the next term is generally determined by applying some arithmetic operations. An arithmetic progression is required in many fields, like physics, engineering, economics, etc., for predicting sequences or analysing trends by observing the rate of change.

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These notes covered all the topics and subtopics of arithmetic progression, like the basic definition of arithmetic progression, finite and infinite arithmetic progression, general term of arithmetic progression, sum of the nth term and arithmetic mean and formulae required to calculate the arithmetic progression. CBSE Class 10 chapter Arithmetic Progression also includes the nth term of the arithmetic progression. NCERT class 10th maths notes are designed by our experts that help the students understand the concepts effectively, and students can download NCERT notes according to their standard and subjects.

NCERT Class 10 Maths Chapter 5 Arithmetic Progressions: Notes

Arithmetic Progression: An Arithmetic progression is a sequence, list or series of numbers where the difference between two terms is constant, and we can get the preceding and succeeding terms by performing the arithmetic operation.
Arithmetic progression is denoted by $AP$, and the first term is denoted by $a$₁ and the last term is denoted by $a$_n. It implies that the number of terms in $AP$ is $n$.
Example: 5, 8, 11, 14, 17, 20, 23......

Common difference: In an arithmetic progression, the difference between two consecutive terms is called the common difference, and it is denoted by $d$.
Example: 7, 12, 17, 22, 27, 32
In this arithmetic progression, the common difference ($d$) = 5

Formula For Calculating Common Difference:
$d$ = ($a$_n $-$ $a$_{n - 1})
Where,
$d$ = Common difference
$a$_n = nth term of the sequence
$a$_{n - 1}) = (n - 1)th term of the sequence
There are three types of common difference $(d)$:

1. Positive Difference: The $AP$ is increasing in this case.
Example: 1, 2, 3, 4, 5

2. Negative Difference: In this case, the $AP$ decreases.
Example: 100, 90, 80, 70, 60, 50

3. Zero Difference: In this case, the $AP$ is constant.
Example: 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3

Types of Arithmetic Progression (AP):

There are two types of arithmetic progression (AP):

1. Finite Arithmetic Progression: An arithmetic progression (AP) in which the number of terms is finite is called a finite arithmetic progression (AP). The finite AP have an nth term.
Example: 1, 3, 5, 7, 9, 11

2. Infinite Arithmetic Progression: An arithmetic progression (AP) in which the number of terms is infinite is called an infinite arithmetic progression (AP). The infinite AP does not have the nth term.
Example: 2, 4, 6, 8, 10, ........

The nth Term Of An Arithmetic Progression (AP):

The nth term in the last term in AP and the formula for determining the nth term is as follows:
$a$_n = $a+(n-1)d$
Here,
a = First term
d = Common difference
n = number of terms

Example: Determine the 10th term of the arithmetic progression 4, 8, 12, 16, .......
Given:
AP = 4, 8, 12, 16, .......
Here, a = 4
(8 - 4) = 4 and (12 - 8) = 4
So, d = 4
And now, we have to find the 10th term of the given AP.
Therefore, n = 10
$a$_n = $a+(n-1)d$
After substituting the values in the formula, we get
a₁₀ = 4 + (10 – 1)4
a₁₀ = 4 + 9 × 4
a₁₀ = 4 + 36
a₁₀ = 40
So, the 10th term of the given arithmetic sequence 4, 8, 12, 16, ....... is 40

The General Form Of An Arithmetic Progression (AP):

The general form of an AP is as follows:
a, a + d, a + 2d, a + 3d, a + 4d ....
Where a is the first term and d is the common difference, the value of d = 0, d > 0 or d < 0.

The Sum Of The Terms In Arithmetic Progression (AP):

The formula for calculating the sum of the terms in AP is as follows:

$S$ = $\frac{n}{2}$ $[2a+(n-1)d]$

Where,
S = Sum of the terms
n = Number of terms
a = First term
d = Common difference
Or we can also write this as,

$S$ = $\frac{n}{2}$ $(a+a$_n)

Example: Determine the sum of the first 18 terms of AP 17, 11, 5, -1, .......
Given:
AP = 17, 11, 5, -1, .......
First term a = 17
Common difference d = 11 - 17 = -6, and 5 - 11 = -6
Therefore, the common difference d = -6
Number of terms n = 18

$S$ = $\frac{n}{2}$ $[2a+(n-1)d]$

After substituting all the values, we get

$S$ = $\frac{18}{2}$ $[2\times 17+(18-1)\times (-6)]$

$S$ = $9$ × $[34+(17)\times (-6)]$

$S$ = $9$ × $[34+(-102)]$

$S$ = $9$ × $(-68)$

$S$ = $-612$
Therefore, the sum of the 18th term is -612.

Arithmetic Mean (AM):

The average of the terms in an arithmetic progression is called the Arithmetic Mean (AM). Therefore, the arithmetic mean formula is as follows:
$AM$ = $\frac{\text{Sum of the terms}}{\text{Number of the terms}}$

The Sum of the First n Natural Numbers:

The sum of the first n natural numbers can be determined by:

$S$_n = $\frac{n(n+1)}{2}$

Here,
n = Number of terms

Example: Determine the sum of the first 15 natural numbers.
Given:
n = 15

$S$_n = $\frac{n(n+1)}{2}$

$S$_n = $\frac{15(15+1)}{2}$

$S$_n = $\frac{15\times 16}{2}$

$S$_n = $120$
Therefore, the sum of the first 15 natural numbers is 120.

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