CBSE Notes Class 10 Maths Chapter 5 Arithmetic Progressions

1.0Introduction to Arithmetic Progressions

Have you ever seen a sequence in your daily life, at work, or in other things? Sequences are like sunflower petals or holes in a honeycomb. This type of sequence in Maths is called Arithmetic progression. For example, let’s say you invest Rs. 5000 in January, and each month, the amount adds up by interest of Rs. 500. This is a sequence where the monthly increment of the principal amount is Rs 500. 

2.0Download CBSE Notes Class 10 Maths Chapter 5: Arithmetic Progressions - Free PDF!!

Unlock a comprehensive understanding of Arithmetic Progressions in CBSE Class 10 with these free Maths Chapter 5 notes in PDF format. Master the concepts and formulas to ace your exams!

3.0CBSE Maths Notes For Class 10 Chapter 5 Arithmetic Progressions - Revision Notes

Definition of Arithmetic Progression (AP)

The Arithmetic Progression is a pattern or sequence of things or numbers where the difference between each component of the sequence remains constant. This constant difference is termed a common difference (d).

Note: The common difference is the defining factor to determine whether a given sequence is an Arithmetic Progression or not. 

The General form of AP 

a, a+d, a+2d, a+3d ……..

d = a_n - a_n-1 

• a = First term. 
• d = Common difference. (The common difference can be negative, positive, or zero.)
• n = 1, 2, 3, 4, ………, n

General Form of n^th Term of Arithmetic Progression 

The formula for the nth term is 

a_n = a + (n - 1)d 

n = no. of terms in a given sequence. 

A_n = nth term or last term of AP. It is also denoted by l. 

Let’s understand AP with an example.

Example: Write the first four terms of an AP whose a = 2 and d = 3 are given below as follows. 

Solution: a₁ = 2 

a₂ = 2 + (2 - 1)3 = 2 + (1)3 = 2 + 3 = 5 

a₃ = 2 + (3 - 1)3 = 2 + (2)3 = 2 + 6 = 8

a₄ = 2 + (4 - 1)3 = 2 + (3)3 = 2 + 9 = 11

Example: What will be the 10^th term of multiple of 5? 

Solution: d = 10 - 5 = 5 

a₁₀ = 5 + (10-1)5 = 5 + (9)5 = 5 + 45 = 50 

Example: Find the last term of an AP 7, 14, 21,…... given that it contains 14 terms. 

Solution: d = 14 - 7 = 7

a₁₄ = 7 + (13-1)7 = 7 + (12)7 = 7 + 84 = 91.

Example: The sum of the 4th and 8th terms of an AP is 32, and the sum of the 6th and 10th terms is 52. Find the first term of the AP.

Solution: we have a₄ = 32 

a + 3d = 32 ……………………1

And a₈ = 52 

a + 7d = 52 ………………………….2

Eliminating equation 1 and 2 

a + 3d = 32

a + 7d = 52

4d = 20 

d = 5 

a + 3 5 = 32

a = 32 - 15 = 17

What if we sum up the terms of AP?

Let us once again consider the example that is given at the start where we get 500 interest every month. What will be the total amount collected after 6 months of investment? The total amount will be 8000. So, what we did here is simply sum up all the interest to the invested amount. 

The sum of n terms of AP is the same process where we sum up all the terms of AP. 

The formula for the Sum of nth terms of an AP 

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

Or alternatively:

${\mathrm{S}}_{\mathrm{n}}=\frac{n}{2}\left(a+a_n\right)$
Where:

• S_n is the sum of the first n terms
• a is the first term
• d is the common difference
• n is the number of terms
• a_n​ is the nth term

Points to note: 

• a₁ = s₁ 
• a_n = S_n - S_n-1 
• The sum of the first n positive integers is n(n-1)/2

Example: Find the sum of the first 20 multiples of 6. 

Solution: n = 20, a = 6, d = 6

$S_{20}=\frac{20}{2}(2\times 6+(20-1)6)=10(12+19\times 6)=1260$

4.0Key Features of CBSE Maths Notes for Class 10 of Chapter 5

• The notes are in line with the latest CBSE curriculum. 
• The language of the notes is easy to understand, making it ideal for self-learning. 
• Every concept of notes is provided with a solved problem to get a better understanding of AP. 
• The notes are made under the guidance of our experienced faculty to maintain accuracy.