# NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progression Exercise 5.1

## NCERT Solutions Class 10 Maths Chapter 5 Arithmetic Progression Ex 5.1 Students can download NCERT Maths Class 10 Chapter 5 Exercise 5.1 PDF from Vedantu for free. The PDF contains solutions to the sums given in the exercise with clearly defined steps for the understanding of students. The familiarity with these concepts can be achieved through practice. Class 10 Chapter 5 Exercise 5.1 introduces the concepts of Arithmetic progression. It is one of the foundational concepts in mathematics. With a firm grasp of the topic and an in-depth understanding of the concepts governing progressions, students can score well in their upcoming exams and also in their higher studies. Our curated solution for CBSE NCERT books for Class 10 Maths has a specific focus on exam preparation. With these Solutions of 10th class maths, students can acquire in-depth knowledge of all the chapters. Candidates can download NCERT solution PDF from Vedantu and continue with their exam preparation.

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Exercise 5.1

1. In which of the following situations, does the list of numbers involved make as arithmetic progression and why?

1. The taxi fare after each km when the fare is Rs $15$ for the first km and Rs $8$ for each additional km.

Ans: Given the fare of first km is Rs.$15$ and the fare for each additional km is Rs. $8$. Hence,

Taxi fare for ${{1}^{st}}$ km is Rs. $15$.

Taxi fare for ${{2}^{nd}}$ km is Rs. $15+8=23$.

Taxi fare for ${{3}^{rd}}$ km is Rs. $23+8=31$.

Similarly, Taxi fare for ${{n}^{th}}$ km is Rs. $15+\left( n-1 \right)8$.

Therefore, we can conclude that the above list forms an A.P with common difference of $8$.

1. The amount of air present in a cylinder when a vacuum pump removes a quarter of the air remaining in the cylinder at a time.

Ans: Let the initial volume of air in a cylinder be $V$ liter. In each stroke, the vacuum pump removes $\dfrac{1}{4}$ of air remaining in the cylinder at a time. Hence,

Volume after ${{1}^{st}}$ stroke is $\dfrac{3V}{4}$.

Volume after ${{2}^{nd}}$ stroke is $\dfrac{3}{4}\left( \dfrac{3V}{4} \right)$.

Volume after ${{3}^{rd}}$ stroke is ${{\left( \dfrac{3}{4} \right)}^{2}}\left( \dfrac{3V}{4} \right)$.

Similarly, Volume after ${{n}^{th}}$ stroke is ${{\left( \dfrac{3}{4} \right)}^{n}}V$.

We can observe that the subsequent terms are not added with a constant digit but are being multiplied by $\dfrac{3}{4}$. Therefore, we can conclude that the above list does not forms an A.P.

1. 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.

Ans: Given the cost of digging for the first meter is Rs.$150$ and the cost for each additional meter is Rs. $50$. Hence,

Cost of digging for ${{1}^{st}}$ meter is Rs. $150$.

Cost of digging for ${{2}^{nd}}$ meter is Rs. $150+50=200$.

Cost of digging for ${{3}^{rd}}$ meter is Rs. $200+50=250$.

Similarly, Cost of digging for ${{n}^{th}}$ meter is Rs. $150+\left( n-1 \right)50$.

Therefore, we can conclude that the above list forms an A.P with common difference of $50$.

1. The amount of money in the account every year, when Rs $\mathbf{10000}$ is deposited at compound interest at $\mathbf{8}\%$ per annum.

Ans: Given the principal amount is Rs.$\mathbf{10000}$ and the compound interest is $\mathbf{8}\%$ per annum. Hence,

Amount after ${{1}^{st}}$ year is Rs. $10000\left( 1+\dfrac{8}{100} \right)$.

Amount after ${{2}^{nd}}$ year is Rs. $10000{{\left( 1+\dfrac{8}{100} \right)}^{2}}$.

Amount after ${{3}^{rd}}$ year is Rs. $10000{{\left( 1+\dfrac{8}{100} \right)}^{3}}$.

Similarly, Amount after ${{n}^{th}}$ year is Rs. $10000{{\left( 1+\dfrac{8}{100} \right)}^{n}}$.

We can observe that the subsequent terms are not added with a constant digit but are being multiplied by $\left( 1+\dfrac{8}{100} \right)$. Therefore, we can conclude that the above list does not forms an A.P.

2. Write first four terms of the A.P. when the first term $a$ and the common difference $d$ are given as follows:

1.  $a=10,d=10$

Ans: We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$       …..(1)

Substituting $a=10,d=10$ in (1) we get, ${{a}_{n}}=10+10\left( n-1 \right)=10n$ …..(2)

Therefore, from (2)

${{a}_{1}}=10$, ${{a}_{2}}=20$, ${{a}_{3}}=30$ and ${{a}_{4}}=40$.

1. $a=-2,d=0$

Ans: We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$       …..(1)

Substituting $a=-2,d=0$ in (1) we get, ${{a}_{n}}=-2+0\left( n-1 \right)=-2$ …..(2)

Therefore, from (2)

${{a}_{1}}=-2$, ${{a}_{2}}=-2$, ${{a}_{3}}=-2$ and ${{a}_{4}}=-2$.

1. $a=4,d=-3$

Ans: We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$       …..(1)

Substituting $a=4,d=-3$ in (1) we get, ${{a}_{n}}=4-3\left( n-1 \right)=7-3n$ …..(2)

Therefore, from (2)

${{a}_{1}}=4$, ${{a}_{2}}=1$, ${{a}_{3}}=-2$ and ${{a}_{4}}=-5$.

1. $a=-1\text{,}d=1/2$

Ans: We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$       …..(1)

Substituting $a=-1\text{,}d=1/2$ in (1) we get, ${{a}_{n}}=-1+\dfrac{1}{2}\left( n-1 \right)=\dfrac{n-3}{2}$ …..(2)

Therefore, from (2)

${{a}_{1}}=-1$, ${{a}_{2}}=-\dfrac{1}{2}$, ${{a}_{3}}=0$ and ${{a}_{4}}=\dfrac{1}{2}$.

1. $a=-1.25,d=-0.25$

Ans: We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$       …..(1)

Substituting $a=-1.25,d=-0.25$ in (1) we get, ${{a}_{n}}=-1.25-0.25\left( n-1 \right)=-1-0.25n$ …..(2)

Therefore, from (2)

${{a}_{1}}=-1.25$, ${{a}_{2}}=-1.5$, ${{a}_{3}}=-1.75$ and ${{a}_{4}}=-2$.

3. For the following A.P.s, write the first term and the common difference.

1. $\mathbf{3},\mathbf{1},-\mathbf{1},-\mathbf{3},...$

Ans: From the given AP, we can see that the first term is $3$.

The common difference is the difference between any two consecutive numbers of the A.P.

Common difference $=$ ${{2}^{nd}}\text{ }term-{{1}^{st}}\ term$

$\therefore$ Common difference $=$ $1-3=-2$.

1. $-5,-1,3,7,...$

Ans: From the given AP, we can see that the first term is $-5$.

The common difference is the difference between any two consecutive numbers of the A.P.

Common difference $=$ ${{2}^{nd}}\text{ }term-{{1}^{st}}\ term$

$\therefore$ Common difference $=$ $-1-\left( -5 \right)=4$.

1. $\dfrac{1}{3}\text{,}\dfrac{5}{3}\text{,}\dfrac{9}{3}\text{,}\dfrac{13}{3},...$

Ans: From the given AP, we can see that the first term is $\dfrac{1}{3}$.

The common difference is the difference between any two consecutive numbers of the A.P.

Common difference $=$ ${{2}^{nd}}\text{ }term-{{1}^{st}}\ term$

$\therefore$ Common difference $=$ $\dfrac{5}{3}-\dfrac{1}{3}=\dfrac{4}{3}$.

1. $0.6,1.7,2.8,3.9,...$

Ans: From the given AP, we can see that the first term is $0.6$.

The common difference is the difference between any two consecutive numbers of the A.P.

Common difference $=$ ${{2}^{nd}}\text{ }term-{{1}^{st}}\ term$

$\therefore$ Common difference $=$ $1.7-0.6=1.1$.

4. Which of the following are AP’s? If they form an AP, find the common difference $d$ and write three more terms.

1.  $\text{2,4,8,16}...$

Ans: For the given series, let us check the difference between all consecutive terms and find if they are equal or not.

${{a}_{2}}-{{a}_{1}}=4-2=2$              …..(1)

${{a}_{3}}-{{a}_{2}}=8-4=4$              …..(2)

${{a}_{4}}-{{a}_{3}}=16-8=8$              …..(3)

From (1), (2), and (3) we can see that the difference between all consecutive terms is not equal.

Therefore, the given series does not form an A.P.

1. $\text{2,}\dfrac{5}{2}\text{,3,}\dfrac{7}{2}\text{,}...$

Ans: For the given series, let us check the difference between all consecutive terms and find if they are equal or not.

${{a}_{2}}-{{a}_{1}}=\dfrac{5}{2}-2=\dfrac{1}{2}$              …..(1)

${{a}_{3}}-{{a}_{2}}=3-\dfrac{5}{2}=\dfrac{1}{2}$              …..(2)

${{a}_{4}}-{{a}_{3}}=\dfrac{7}{2}-3=\dfrac{1}{2}$              …..(3)

From (1), (2), and (3) we can see that the difference between all consecutive terms is equal.

Therefore, the given series form an A.P. with first term $2$ and common difference $\dfrac{1}{2}$.

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$       …..(4)

Substituting $a=2,d=\dfrac{1}{2}$ in (1) we get, ${{a}_{n}}=2+\dfrac{1}{2}\left( n-1 \right)=\dfrac{n+3}{2}$ …..(5)

Therefore, from (5)

${{a}_{5}}=4$, ${{a}_{6}}=\dfrac{9}{2}$ and ${{a}_{7}}=5$.

1. $1.2,3.2,5.2,7.2...$

Ans: For the given series, let us check the difference between all consecutive terms and find if they are equal or not.

${{a}_{2}}-{{a}_{1}}=3.2-1.2=2$              …..(1)

${{a}_{3}}-{{a}_{2}}=5.2-3.2=2$              …..(2)

${{a}_{4}}-{{a}_{3}}=7.2-5.2=2$              …..(3)

From (1), (2), and (3) we can see that the difference between all consecutive terms is equal.

Therefore, the given series form an A.P. with first term $1.2$ and common difference $2$.

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$       …..(4)

Substituting $a=1.2,d=2$ in (1) we get, ${{a}_{n}}=1.2+2\left( n-1 \right)=2n-0.8$ …..(5)

Therefore, from (5)

${{a}_{5}}=9.2$, ${{a}_{6}}=11.2$ and ${{a}_{7}}=13.2$.

1. $-10,-6,-2,2,...$

Ans: For the given series, let us check the difference between all consecutive terms and find if they are equal or not.

${{a}_{2}}-{{a}_{1}}=-6-\left( -10 \right)=4$              …..(1)

${{a}_{3}}-{{a}_{2}}=-2-\left( -6 \right)=4$              …..(2)

${{a}_{4}}-{{a}_{3}}=2-\left( -2 \right)=4$              …..(3)

From (1), (2), and (3) we can see that the difference between all consecutive terms is equal.

Therefore, the given series form an A.P. with first term $-10$ and common difference $4$.

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$       …..(4)

Substituting $a=-10,d=4$ in (1) we get, ${{a}_{n}}=-10+4\left( n-1 \right)=4n-14$ …..(5)

Therefore, from (5)

${{a}_{5}}=6$, ${{a}_{6}}=10$ and ${{a}_{7}}=14$.

1. $3,3+\sqrt{2},3+2\sqrt{2},3+3\sqrt{2},...$

Ans: For the given series, let us check the difference between all consecutive terms and find if they are equal or not.

${{a}_{2}}-{{a}_{1}}=\left( 3+\sqrt{2} \right)-\left( 3 \right)=\sqrt{2}$              …..(1)

${{a}_{3}}-{{a}_{2}}=\left( 3+2\sqrt{2} \right)-\left( 3+\sqrt{2} \right)=\sqrt{2}$              …..(2)

${{a}_{4}}-{{a}_{3}}=\left( 3+3\sqrt{2} \right)-\left( 3+2\sqrt{2} \right)=\sqrt{2}$              …..(3)

From (1), (2), and (3) we can see that the difference between all consecutive terms is equal.

Therefore, the given series form an A.P. with first term $3$ and common difference $\sqrt{2}$.

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$       …..(4)

Substituting $a=3,d=\sqrt{2}$ in (1) we get, ${{a}_{n}}=3+\left( n-1 \right)\sqrt{2}$   …..(5)

Therefore, from (5)

${{a}_{5}}=3+4\sqrt{2}$, ${{a}_{6}}=3+5\sqrt{2}$ and ${{a}_{7}}=3+6\sqrt{2}$.

1. $\text{0}\text{.2,0}\text{.22,0}\text{.222,0}\text{.2222}.....$

Ans: For the given series, let us check the difference between all consecutive terms and find if they are equal or not.

${{a}_{2}}-{{a}_{1}}=0.22-0.2=0.02$              …..(1)

${{a}_{3}}-{{a}_{2}}=0.222-0.22=0.002$              …..(2)

${{a}_{4}}-{{a}_{3}}=0.2222-0.222=0.0002$              …..(3)

From (1), (2), and (3) we can see that the difference between all consecutive terms is not equal.

Therefore, the given series does not form an A.P.

1. $0,-4,-8,-12....$

Ans: For the given series, let us check the difference between all consecutive terms and find if they are equal or not.

${{a}_{2}}-{{a}_{1}}=-4-0=-4$              …..(1)

${{a}_{3}}-{{a}_{2}}=-8-\left( -4 \right)=-4$              …..(2)

${{a}_{4}}-{{a}_{3}}=-12-\left( -8 \right)=-4$              …..(3)

From (1), (2), and (3) we can see that the difference between all consecutive terms is equal.

Therefore, the given series form an A.P. with first term $0$ and common difference $-4$.

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$       …..(4)

Substituting $a=0,d=-4$ in (1) we get, ${{a}_{n}}=0-4\left( n-1 \right)=4-4n$ …..(5)

Therefore, from (5)

${{a}_{5}}=-16$, ${{a}_{6}}=-20$ and ${{a}_{7}}=-24$.

1. $-\dfrac{1}{2},-\dfrac{1}{2},-\dfrac{1}{2},-\dfrac{1}{2}....$

Ans: For the given series, let us check the difference between all consecutive terms and find if they are equal or not.

${{a}_{2}}-{{a}_{1}}=\left( -\dfrac{1}{2} \right)-\left( -\dfrac{1}{2} \right)=0$              …..(1)

${{a}_{3}}-{{a}_{2}}=\left( -\dfrac{1}{2} \right)-\left( -\dfrac{1}{2} \right)=0$              …..(2)

${{a}_{4}}-{{a}_{3}}=\left( -\dfrac{1}{2} \right)-\left( -\dfrac{1}{2} \right)=0$              …..(3)

From (1), (2), and (3) we can see that the difference between all consecutive terms is equal.

Therefore, the given series form an A.P. with first term $-\dfrac{1}{2}$ and common difference $0$.

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$       …..(4)

Substituting $a=-\dfrac{1}{2},d=0$ in (1) we get, ${{a}_{n}}=-\dfrac{1}{2}+0\left( n-1 \right)=-\dfrac{1}{2}$ …..(5)

Therefore, from (5)

${{a}_{5}}=-\dfrac{1}{2}$, ${{a}_{6}}=-\dfrac{1}{2}$ and ${{a}_{7}}=-\dfrac{1}{2}$.

## NCERT Solutions for Class 10 Maths - Exercise 5.1

#### NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.1 Free PDF Download

Class 10 maths ex 5.1 solutions provided in the PDF are written by our subject experts at Vedantu. These are written in accordance with the CBSE guidelines so students can refer to them and take notes as to how they should format their answers in their examination. The maths paper follows step-wise marking, where you are awarded marks for every step, so it is essential to understand how to write answers correctly.

The class 10th maths chapter 5 exercise 5.1 pdf is written in a simple language to maximize retention. The focus is on increasing the concept-based understanding of students, which will lead you to score well in your exams. Your class 10th score matters a lot as students are distributed streams according to their class 10 score.

Prepare for your exams better with class 10 maths ex 5.1 solutions. Students are advised to cultivate a habit of practicing questions from the NCERT book regularly as the questions in your exams will be similar to the ones asked in your NCERT book. Download exercise 5.1 class 10 maths NCERT solutions here.

Now, take a look at other exercises in this chapter.

### Class 10 Maths Weightage Marks

Class 10 maths paper is divided into four sections- Section A, B, C, and D. The paper is designed to test both mathematical and application-based knowledge in students. The types of questions students will find in their question paper are:

• Very Short Answer Type Questions - 20 questions x 1 mark each.

• Short Answer Type I Questions - 6 questions x 2 marks each.

• Short Answer Type II Questions - 8 questions x 3 marks each.

• Long Answer Type Questions - 6 questions x 4 marks each.

The class 10th maths syllabus is diversified so as to allow students to form a firm foundation for important mathematical concepts that will help them in their further studies. Each chapter is interconnected and requires constant practice to be able to score well. Give below are the weightage marks of each unit so that students can plan their preparations effectively and score well in their exams.

 Unit Weightage Marks Number System 06 Algebra 20 Coordinate Geometry 06 Geometry 15 Trigonometry 12 Mensuration 10 Statistics And Probability 11

### Benefits of Ex 5.1 Class 10 Maths Solutions

The chapter arithmetic progression is governed by some simple concepts. If you understand these concepts well, it is mostly formula-based. The formulas for the nth term and sum of an AP should be on your tips. Students will be required to convert word problems into mathematical equations. This can be tricky but with correct guidance from NCERT Class 10 Maths Chapter 5 exercise 5.1 solutions PDF, it will be smooth sailing.

Students can freely download NCERT solutions for class 10 maths chapter 5 exercise 5.1 here. The benefits of using solutions of class 10 arithmetic progression exercise 5.1 pdf are:

• The notes are prepared in simple language to improve understanding of the concepts and steps.

• These solutions are written in accordance with CBSE guidelines to help you score maximum marks in the examinations.

• The solutions pdf is prepared with the help of experienced teachers.

• Comprehensive and detailed explanations of each question allow students to understand the topic and the concept well.

Question 1. What are Some of the Important Concepts of the Chapter Arithmetic Progression?

Ans: Arithmetic progression is a sequence of numbers where the difference between two consecutive numbers is the same. This difference between two consecutive terms is called the common difference. As long as the sequence or series of numbers follow this rule they are said to be in an AP.

An AP can be commonly written as:

a, a + d, a + 2d, a + 3d, a + 4d, ......, a + (n - 1)d.

If the first term a and the common difference d of an AP is known we can find the value of the nth or any term of an AP  easily.

an = a + (n - 1)d

The sum of the AP can be calculated using the formula:

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

Question 2. How can the Sum of an AP be Calculated? Derive the Formula Mathematically for the Same.

Ans: The sum of an AP is the sum of each term of the AP. The summation of each term of an AP can be tiresome in longer progressions. Also, the APs with a negative common difference, finding the sum of AP can be tricky. The sum of the AP can be calculated using the following formula:

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

Derivation:

Sn = a1 + (a1 + d) + (a1 + 2d) + .... (a1 + (n - 1)d). (i)

Sn = an + (an - d) + (an - 2d) + ...(an - (n - 1)d). (ii)

Adding both the equations and solving,

2Sn = n(a1 + an)

an = a1 + (n - 1)d

2Sn = n(a1 + a1 + (n - 1)d)

2Sn = n(2a1 + (n - 1)d)

S = n/2(2a1 + (n - 1)d)

The formula for sum of AP can be written as:

S = n/2(2a + (n - 1)d) SHARE TWEET SHARE SUBSCRIBE