Class 10 Exam  >  Class 10 Notes  >  Mathematics (Maths) Class 10  >  Chapter Notes: Arithmetic Progressions

Arithmetic Progressions Class 10 Notes Maths Chapter 5

Introduction

Patterns can be observed in our daily lives, even in something as simple as a savings account. For example, if we start with Rs. 2000 and add Rs. 500 each month, the balance follows a predictable pattern: Rs. 2000, Rs. 2500, Rs. 3000, and so on. This is called an Arithmetic progression (AP).

Arithmetic Progressions

Arithmetic Progressions (APs) are sequences of numbers in which the difference between consecutive terms remains constant. This constant difference is called the common difference. In an arithmetic progression, each term can be obtained by adding the common difference to the previous term.

Here is a list of numbers, observe the pattern of the numbers.
Arithmetic Progressions Class 10 Notes Maths Chapter 5

  • Here, we see that each successive term is obtained by adding a fixed number to the preceding term except the first term. Such a list of numbers is said to be in arithmetic progression.
  • This fixed number is called the common difference of the AP. This number can be positive, negative, or zero.
  • Let us denote the first term of an AP by a1, the second term by a2,.........nth term by an and the common difference by d. Then the AP becomes a1, a2, a3, ... ... ... ... ... aₙ.
    So, a1 − a2 = a3 − a2 =.........= an − an−1 = d

Arithmetic Progressions (APs) can be categorized as either finite or infinite based on the number of terms they contain.

  • Finite Arithmetic Progression: A finite Arithmetic Progression (AP) refers to a sequence of numbers that has a specific and limited number of terms. The progression stops after a certain term, and there is a finite number of terms in the sequence.
    Example: 229, 329, 429, 529, 629
  • Infinite Arithmetic Progression: In contrast, an infinite Arithmetic Progression (AP) is a sequence of numbers that continues indefinitely without an endpoint. It goes on infinitely, and there is no fixed limit on the number of terms.
    Example: 2, 4, 6, 8, 10, 12, 14, 16, 18…..…

Understanding the distinction between finite and infinite APs helps us analyze different types of sequences and apply appropriate methods and formulas accordingly.
Arithmetic Progressions Class 10 Notes Maths Chapter 5

Question for Chapter Notes: Arithmetic Progressions
Try yourself:
Which type of arithmetic progression has a fixed and limited number of terms?
View Solution

Solved Examples

Example 1: In which of the following situations, does the list of numbers involved make an arithmetic progression, and why? 

a) The cost of digging a well after every meter of digging when it costs Rs. 150 for the first meter and rises by Rs. 50 for each subsequent meter.

Sol: 

Cost of digging the first meter = Rs.150
Cost of digging second metre = Rs.(150 + 50) = Rs.200
Cost of digging third metre = Rs.(150 + 2× 50) = Rs. 250
In this case, each term is obtained by adding Rs. 50 to the preceding term. Hence, they make an AP.

b) The amount of water present in a cylinder when a vacuum pump removes 1⁄4 of the air remaining in the cylinder at a time.

Sol:

Let the amount of air present in the cylinder be x units.
According to the question,
Amount of air left in the cylinder after using vacuum pump first timeArithmetic Progressions Class 10 Notes Maths Chapter 5
Amount of air left in the cylinder after using vacuum pump the second time=Arithmetic Progressions Class 10 Notes Maths Chapter 5
List of numbers  Arithmetic Progressions Class 10 Notes Maths Chapter 5
Arithmetic Progressions Class 10 Notes Maths Chapter 5
As the common difference of the terms is not the same, they do not form AP.

Example 2: Find the common difference of the AP

Arithmetic Progressions Class 10 Notes Maths Chapter 5

Sol:
Common difference (d) = Second Term – First Term
Arithmetic Progressions Class 10 Notes Maths Chapter 5
Therefore, common difference = -2

Example 3: For what value of k will k + 10 , 2k, and 2k + 8 are the consecutive terms of an AP.

Sol:

If k + 10 , 2k, and 2k + 8 are in AP then,
a1 = k + 10 , a2 = 2k and a3 = 2k + 8

Common Difference (d)
Arithmetic Progressions Class 10 Notes Maths Chapter 5

nth term of an AP

Let a1 , a2, a3, ... ... ... ... ... an be an AP whose first term a1 is a and the common difference is d.

Then,

Arithmetic Progressions Class 10 Notes Maths Chapter 5The nth term an of the AP with the first term a and common difference d is given by 

an = a + (n − 1)d

Question for Chapter Notes: Arithmetic Progressions
Try yourself:
In which of the following situations does the list of numbers involved make an arithmetic progression (AP), and why?

a) The temperature at 9 AM is 15?C, and it decreases by 2?C every hour.
View Solution

Solved Examples

Example 5: Find the 25th term of the AP: -5, -5/2, 0, 5/2

Sol:

Here, a = −5
Common Difference (d) = a2 − a1
Arithmetic Progressions Class 10 Notes Maths Chapter 5

We know, a= a + (n − 1)d

25th term, a25 
Arithmetic Progressions Class 10 Notes Maths Chapter 5
25th term of the given AP is 55.

Example 6: For an AP, if a18 − a14 = 36, then find the common difference d.

Sol:

We know, an = a + (n − 1)d
a18 = a + (18 − 1)d = a + 17d
a14 = a + (14 − 1)d = a + 13d
a18 − a14 = 36
a + 17d − (a + 13d) ⇒ a + 17d − a − 13d
4d = 36
d = 9
Therefore, the common difference (d) = 9

Example 7: If an = 6 − 11n, then find the common difference.

Sol:

Given: a= 6 − 11n ............... (1)

Replacing n by n+1 in eq (1) we get

a(n+1) = 6 – 11(n+1)

a(n+1) = 6 – 11n-11

a(n+1) = - 11n – 5

Common difference, d = a(n+1) – an

d= -11n-5 – (6-11n)

d = -11

Example 8: Find the 7th term of the sequence whose nth term is given by an = (−1)n−1. n2

Sol:

Given: an = (−1)n−1. n2
The 7th term of the sequence, a7 = (−1)7−1. 72
= (−1)6. 72 = 49

Example 9: An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.

Sol:

Let a be the first term and d be the common difference of the given AP.

Given: a= 12 and a50 = 106

We know, an = a + (n − 1)d
a3 = a + (3 − 1)d ⇒ a3 = a + 2d
a= a + 2d = 12→Eq 1
a50 = a + (50 − 1)d ⇒ a50 = a + 49d
a50 = a + 49d = 106→Eq 2
Subtracting Eq 1 from Eq 2

a + 49d − (a + 2d) = 106 − 12

a + 49d − a − 2d = 94

47d = 94 ⇒ d = 2

Putting the value of d in Eq 1 we get,

a + 2 × 2 = 12

a = 12 − 4 = 8

29th term, a29 = a + 28d = 8 + 28 × 2 = 8 + 56 = 64

Example 10: Find how many two-digit numbers are divisible by 7.

Sol:

Two-digit numbers are 10, 11, 12, 13, ..................97, 98, 99, 100.
Here 14, 21, 28...................... 91, 98 are divisible by 7.
This list of numbers forms an AP, where a = 14 and
d = 21 − 14 = 7
Let the number of terms be n, then an = 98

98 = 14 + (n − 1)7 ⇒ 98 = 14 + 7n − 7

7n + 7 = 98 ⇒ 7n = 98 − 7

7n = 91

n = 13
Hence, 13 two-digit numbers are divisible by 7.

Question for Chapter Notes: Arithmetic Progressions
Try yourself:In an arithmetic progression (AP), if the first term is 2 and the common difference is 4, what is the 7th term of the AP?
View Solution

Sum of first n terms of an  AP

  • One of the most important aspects of AP is finding the sum of its terms. The sum of the first n terms of an AP is a commonly used formula in mathematics. It allows us to quickly find the total sum of a given number of terms in an AP without having to add each term individually.
  • The sum of first n terms of an AP is given by
    Arithmetic Progressions Class 10 Notes Maths Chapter 5
  • Now, this can also be written as
    Arithmetic Progressions Class 10 Notes Maths Chapter 5
  • We know an = a + (n − 1)d
    Arithmetic Progressions Class 10 Notes Maths Chapter 5
  • If  the number of terms in the AP is n then, an is the last term and an = I
    therefore, Arithmetic Progressions Class 10 Notes Maths Chapter 5
  • This form of the result is useful when only the first and the last term are given and the common difference is not given.

Question for Chapter Notes: Arithmetic Progressions
Try yourself:For an arithmetic progression (AP), if the first term is 10 and the common difference is 3, what is the 15th term?
View Solution

Example 11: If the nth term of an AP is (2n + 2), find the sum of first n terms of the AP.

We have,
an = 2n + 2 ⇒ a1 = 2 × 1 + 2 = 4
Therefore, a= a is the first term and an = l is the last term of the AP.
As we know the first and the last term of the AP, the sum of n terms is given by,
Arithmetic Progressions Class 10 Notes Maths Chapter 5
Arithmetic Progressions Class 10 Notes Maths Chapter 5
Sum of first n terms of the given AP is n[n + 3]

Example 12: Find the sum of the first 25 terms of an AP, whose nth term is given by an = 6 − 3n.

Given: an = 6 − 3n
a1 = 6 − 3 × 1 = 3

a25 = 6 − 3 × 25 = −69
Therefore, a1 = a = 3 is the first term and a25 = l = −69
Arithmetic Progressions Class 10 Notes Maths Chapter 5
Therefore, the sum of the first 25 terms of the given AP is -825.

Example 13: Find the sum of all two-digit odd positive numbers.

Two-digit odd positive numbers are 11, 13, 15..................99 which form an AP.
Here, First term a = 11, last term (l) = 99 and common difference (d) = 13 – 11 = 2
Now we have to find the number of terms.

We know, l = an = a + (n − 1)d
99 = 11 + (n − 1) × 2
99 = 11 + 2n − 2 ⇒ 99 = 9 + 2n
99 − 9 = 2n ⇒ 90 = 2n
Arithmetic Progressions Class 10 Notes Maths Chapter 5
Therefore, the sum of all two-digit odd positive numbers is 2475.

The document Arithmetic Progressions Class 10 Notes Maths Chapter 5 is a part of the Class 10 Course Mathematics (Maths) Class 10.
All you need of Class 10 at this link: Class 10
124 videos|457 docs|77 tests

Top Courses for Class 10

FAQs on Arithmetic Progressions Class 10 Notes Maths Chapter 5

1. What is the formula to find the nth term of an Arithmetic Progression (AP)?
Ans. The formula to find the nth term of an AP is: \(a_n = a_1 + (n-1)d\), where \(a_n\) is the nth term, \(a_1\) is the first term, \(n\) is the term number, and \(d\) is the common difference.
2. How can you determine the sum of the first n terms of an Arithmetic Progression (AP)?
Ans. The formula to find the sum of the first n terms of an AP is: \(S_n = \frac{n}{2}(2a_1 + (n-1)d)\), where \(S_n\) is the sum of the first n terms, \(a_1\) is the first term, \(n\) is the number of terms, and \(d\) is the common difference.
3. How can you identify if a given sequence is an Arithmetic Progression (AP)?
Ans. To identify if a given sequence is an AP, check if the difference between consecutive terms is constant. If the difference is the same throughout the sequence, then it is an Arithmetic Progression.
4. Can an Arithmetic Progression have a common difference of zero?
Ans. Yes, an Arithmetic Progression can have a common difference of zero. In this case, all the terms in the sequence will be the same, forming a constant sequence.
5. How can you find the number of terms in an Arithmetic Progression when given the first term, last term, and common difference?
Ans. To find the number of terms in an AP when given the first term, last term, and common difference, you can use the formula: \(n = \frac{l - a_1}{d} + 1\), where \(n\) is the number of terms, \(l\) is the last term, \(a_1\) is the first term, and \(d\) is the common difference.
124 videos|457 docs|77 tests
Download as PDF
Explore Courses for Class 10 exam

Top Courses for Class 10

Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev
Related Searches

shortcuts and tricks

,

Free

,

video lectures

,

Extra Questions

,

Viva Questions

,

mock tests for examination

,

Summary

,

Sample Paper

,

ppt

,

pdf

,

Arithmetic Progressions Class 10 Notes Maths Chapter 5

,

Arithmetic Progressions Class 10 Notes Maths Chapter 5

,

MCQs

,

Exam

,

Important questions

,

practice quizzes

,

Semester Notes

,

Objective type Questions

,

Previous Year Questions with Solutions

,

study material

,

Arithmetic Progressions Class 10 Notes Maths Chapter 5

,

past year papers

;