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NCERT Solutions For Class 10 Maths Chapter 5 Arithmetic Progressions

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NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions - Free PDF

NCERT Solutions Maths Class 10 Chapter 5 Arithmetic Progressions are provided by Vedantu. Some of the experts at Vedantu have years of experience and knowledge and they have prepared the chapter 5 Maths Class 10 solutions. Their number of years of working experience and knowledge might help students in solving the questions of chapter 5 Maths Class 10 solutions. NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions pdf are now available for all the students to download and get everything they want in one place only. NCERT Solutions for each subject and class are available for students to download. Science students can download Class 10 Science Solutions which has been curated by master teachers at Vedantu. 

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions part-1
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FAQ (Frequently Asked Questions)

1. What is the formula for finding the nth term of an AP? In the formula for the nth term of an AP, can the value of ‘n’, ‘a’ and ‘d’ be a negative integer?

The formula for finding the nth term of an AP is aₙ = a + (n-1)d. In this formula, ‘a’ is the initial term in the arithmetic sequence and ‘d’ is a common difference. These two values can be either a positive or a negative integer. This is because the initial value of the sequence may be a negative value and the difference between any two numbers can also be a negative integer. However, ‘n’ is the number of terms in the sequence. We never use negative integers to count anything. Negative numbers are not counting numbers. Counting begins with the number ‘1’ and increases gradually in the increments of one. All the counting numbers are natural numbers. So, the value of ‘n’ in this formula can never be a negative integer.

2. How does the NCERT Solutions for Class 10 Maths Chapter 5 pdf developed by Vedantu help students in improving their scores?

NCERT Solution for Class 10 Chapter 5 pdf developed by Vedantu Math is unique and accurate. The experienced math experts of Vedantu have always considered the difficulties faced by the adolescent students community and have been successfully delivering student friendly, easily accessible NCERT Solutions for Class 10 Maths Chapter 5 pdf. Each and every question in the exercise is provided with step by step explanation and a crisp recap of the concept in the question. The solutions are preceded by all important formulas in the chapter and also the brief summary with concept map for the entire chapter. This helps the students for the quick revision on a day before their CBSE Board exams at their fingertips.

3. How do you find the sum of first ‘n’ natural numbers, first ‘n’ odd numbers and first ‘n’ even numbers using the concept of arithmetic progression Class 10?

The sum of the terms in an arithmetic progression with ‘n’ terms, ‘a’ as the first term and common difference ‘d’ is given as Sₙ = n/2 [2a + (n-1)d]

The set of natural numbers is given as N = {1, 2, 3, 4, 5 …… n}

The elements in set N are in arithmetic progression because it has a common difference between the consecutive terms. The initial value of the sequence is ‘1’ and the common difference is ‘n’. So, we have to substitute a = 1 and d = 1in the equation for finding the sum of ‘n’ terms of an AP. 

Sₙ = n/2[ 2(1) + (n-1)(1)]

Sₙ = n/2[2 + 2n - 1]

Sₙ = n(n + 1) / 2

The set of ‘n’ odd numbers is written as A = {1, 3, 5, 7, ….. (2n - 1)}

Here a = 1, d = 2. Sum of these terms is given as 

Sₙ = n/2[ 2(1) + (n-1)(2)]

Sₙ = n/2[2 + 2n - 2]

Sₙ = n/2[2n]

Sₙ = n²

The set of ‘n’ even numbers is represented as B = {2, 4, 6, ……. 2n}

In this set a = 2 and d = 2. So the sum of the first ‘n’ even numbers is calculated as:

Sₙ = n/2[ 2(2) + (n-1)(2)]

Sₙ = n/2[4 + 2n - 2]

Sₙ = n/2[2n + 2]

Sₙ = n(n + 1)


4: Can the concept of arithmetic progressions be used in real time examples?

Yes, the concept of arithmetic progressions can explain various real life calculations. Let us take an example to illustrate the application of arithmetic progression formulas in real life. Suhas joined an MNC in the month of April 2020. Suppose he has agreed to work in the same company for the next 10 years with certain payroll norms. The company had assured him a 10% hike on the salary fixed during the joining norms. Let us find his final salary in the month of April 2030. To calculate this, we can use the concept of the nth term of an arithmetic progression. If we assume that his salary is Rs. 12000/- pm when he joined, then the increments of his salary can be represented as:

Salary 1st year = 12,000/- pm

Salary 2nd year = 12,000 + 1,200 = 13,200/- pm

Salary 3rd year = 13,200 + 1200 = 13,400/- pm

So, the salary in the subsequent years per month can be represented in the form of an AP with the first term equal to salary in the first year (12,000), common difference equal to 1,200 and number of terms equal to the number of years (i.e. 10). So, the salary in 10th year can be calculated using the formula for nth term of an AP as:

aₙ = a + (n-1)d

aₙ = 12000 + (10-1)1200

aₙ = 12000 + 10800

aₙ = 22800

The salary received by Sumanth in the 10th year is Rs. 22,800.

5. What is the difference between a sequence and an arithmetic progression?

A sequence is a set of numbers which follow a particular pattern. Each element of a sequence is called its terms. 

For example: 2, 3, 5, 7, ….. is a sequence of Prime numbers.

                      1, 4, 9, 16, ….. is a sequence of squares of natural numbers.

There is no general formula for finding the nth term of a sequence.

Whereas,

An arithmetic progression is a sequence of numbers in which each term is obtained by adding a fixed number to the preceding term except the first term.

For example: 3, 5, 7, 9, …..is an arithmetic progression in which 2 is added to the preceding term to obtain next term.

The nth term of an arithmetic progression is given by: aₙ= a + (n-1) d ; where, an is nth term, a is the first term and d  is common difference of AP.

6. How will you obtain the nth term, if the sum of first n terms and the sum of first (n-1) terms of an arithmetic progression are given?

Given, the sum of first n terms, Sₙ = n/2 [2a + (n-1)d] and

The sum of first (n -1) terms, Sₙ₋₁ = (n-1) / 2 [2a + ((n-1) - 1)d) ….. (obtained by replacing n  with n -1)

The nth term of AP i.e. an can be obtained by subtracting Sₙ and Sₙ₋₁.

So,  subtracting Sₙ₋₁ from Sₙ , we get:

Sₙ - Sₙ₋₁

= n/2 (2a + (n-1)d) - (n-1) / 2 (2a + ((n-1) - 1)d)

= n/2(2a) + [n(n-1)/2 (d)] - [(n-1)/2 (2a)] - [(n-1)(n-2)/2 d].

= na + [n(n-1)/2 (d)] - na + a - [(n-1)(n-2)/2 d]

= [n(n-1)/2 (d)] - na + a - [(n-1)(n-2)/2 d]  …..(obtained after cancelling out na terms)

= a + [(n-1)d]/2 (n-(n-2))

= a + [(n-1)d]/2 (2)

= a + (n-1) d

= aₙ

Therefore, aₙ = Sₙ - Sₙ₋₁


7. The sum of the first n terms of an AP is given by an expression 4n – n². Find its nth term?

Given,  the sum of first n terms of an AP is Sn = 4n –n2

So, S0= 4(0) –(0)2= 0 - 0 = 0

And, S1= 4(1) –(1)2= 4 - 1 = 3

Since we know that: aₙ = Sₙ - Sₙ₋₁

⇒a₁ = S₁ - S₁₋₁

          = S₁ - S₀

          = 3 - 0

⇒a₁ = 3.


Similarly, S2= 4(2) –(2)2= 8 - 4 = 4

⇒a₂ = S₂ - S₂₋₁                ₂₁₋

         = S₂ - S₁

         = 4 - 3

⇒a₂ = 1.

Thus, we have obtained two consecutive terms a1= 3 and a2= 1, it will help us to obtain the common difference of AP.

Common difference = d = a₂ - a₁

                                d = 1 - 3 

                                d = -2

Now, the nth term of an arithmetic progression is given by:

 aₙ = a + (n-1) d 

 where, aₙ is nth term, a is the first term and d  is common difference of AP.

Putting the respective values in above formula, we get:

 aₙ = a + (n-1) d 

          = 3 + (n-1) (-2) 

 = 5 - 2n 

Therefore, the nth term, aₙ = (5 - 2n).

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