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# NCERT Solutions For Class 10 Maths Chapter 5 Exercise 5.1 - Arithmetic Progressions

Last updated date: 07th Sep 2024
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## NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.1 - FREE PDF Download

NCERT Maths Class 10 Chapter 5 Exercise 5.1 PDF from Vedantu is free to download. Ex 5.1 Class 10 Chapter 5 Maths, titled "Arithmetic Progressions," introduces students to sequences and series, specifically focusing on arithmetic progressions (AP). This chapter is crucial for developing a strong foundation in understanding patterns and sequences, which are widely applicable in various mathematical contexts and real-life situations. Our curated solution for CBSE NCERT books for Class 10 Maths has a specific focus on exam preparation. With this 10th Maths Exercise 5.1, students can acquire in-depth knowledge of all the chapters.

Table of Content
1. NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.1 - FREE PDF Download
2. Glance on NCERT Solutions Maths Chapter 5 Exercise 5.1 Class 10 | Vedantu
3. Access NCERT Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Exercise 5.1
4. Class 10 Maths Chapter 5: Exercises Breakdown
5. CBSE Class 10 Maths Chapter 5 Other Study Materials
6. Chapter-Specific NCERT Solutions for Class 10 Maths
FAQs

## Glance on NCERT Solutions Maths Chapter 5 Exercise 5.1 Class 10 | Vedantu

• Exercise 5.1 in Class 10 Maths Chapter 5 deals with Arithmetic Progressions (AP). It likely focuses on understanding the concept and applying basic formulas rather than introducing new ones.

• Find the missing terms in an AP given some terms and the common difference.

• Write the first few terms of an AP given the first term and common difference.

• Find the nth term of an AP.

• Find the sum of the first n terms of an AP.

### Formulas Used

• nth term (general term): $a_n = a + (n - 1)d$

• Sum of n terms: $S_n = \dfrac{n}{2} (a + l)$

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NCERT Solutions For Class 10 Maths Chapter 5 Exercise 5.1 - Arithmetic Progressions
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## Access NCERT Solutions Class 10 Maths Chapter 5 Arithmetic Progressions 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 ${{a}_{n}}=a+\left( n-1 \right)d$       โฆ..(1)

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

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

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

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

Substituting $a=-1.25,d=-0.25$ in (1) we get, ${{a}_{n}}=-1.25-0.25\left( n-1 \right)=-1-0.25n$ โฆ..(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 APโ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.

${{a}_{2}}-{{a}_{1}}=4-2=2$              โฆ..(1)

${{a}_{3}}-{{a}_{2}}=8-4=4$              โฆ..(2)

${{a}_{4}}-{{a}_{3}}=16-8=8$              โฆ..(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.

${{a}_{2}}-{{a}_{1}}=\dfrac{5}{2}-2=\dfrac{1}{2}$              โฆ..(1)

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

${{a}_{4}}-{{a}_{3}}=\dfrac{7}{2}-3=\dfrac{1}{2}$              โฆ..(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 ${{a}_{n}}=a+\left( n-1 \right)d$       โฆ..(4)

Substituting $a=2,d=\dfrac{1}{2}$ in (4) we get, ${{a}_{n}}=2+\dfrac{1}{2}\left( n-1 \right)=\dfrac{n+3}{2}$ โฆ..(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.

${{a}_{2}}-{{a}_{1}}=3.2-1.2=2$              โฆ..(1)

${{a}_{3}}-{{a}_{2}}=5.2-3.2=2$              โฆ..(2)

${{a}_{4}}-{{a}_{3}}=7.2-5.2=2$              โฆ..(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 ${{a}_{n}}=a+\left( n-1 \right)d$       โฆ..(4)

Substituting $a=1.2,d=2$ in (4) we get, ${{a}_{n}}=1.2+2\left( n-1 \right)=2n-0.8$ โฆ..(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.

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

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

${{a}_{4}}-{{a}_{3}}=2-\left( -2 \right)=4$              โฆ..(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 ${{a}_{n}}=a+\left( n-1 \right)d$       โฆ..(4)

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

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

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

${{a}_{4}}-{{a}_{3}}=\left( 3+3\sqrt{2} \right)-\left( 3+2\sqrt{2} \right)=\sqrt{2}$              โฆ..(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 ${{a}_{n}}=a+\left( n-1 \right)d$       โฆ..(4)

Substituting $a=3,d=\sqrt{2}$ in (4) we get, ${{a}_{n}}=3+\left( n-1 \right)\sqrt{2}$   โฆ..(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.

${{a}_{2}}-{{a}_{1}}=0.22-0.2=0.02$              โฆ..(1)

${{a}_{3}}-{{a}_{2}}=0.222-0.22=0.002$              โฆ..(2)

${{a}_{4}}-{{a}_{3}}=0.2222-0.222=0.0002$              โฆ..(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.

${{a}_{2}}-{{a}_{1}}=-4-0=-4$              โฆ..(1)

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

${{a}_{4}}-{{a}_{3}}=-12-\left( -8 \right)=-4$              โฆ..(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 ${{a}_{n}}=a+\left( n-1 \right)d$       โฆ..(4)

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

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

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

${{a}_{4}}-{{a}_{3}}=\left( -\dfrac{1}{2} \right)-\left( -\dfrac{1}{2} \right)=0$              โฆ..(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 ${{a}_{n}}=a+\left( n-1 \right)d$       โฆ..(4)

Substituting $a=-\dfrac{1}{2},d=0$ in (4) we get, ${{a}_{n}}=-\dfrac{1}{2}+0\left( n-1 \right)=-\dfrac{1}{2}$ โฆ..(5)

Therefore, from (5)

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

ix. $1,3,9,27, \ldots$

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

${a_2} - {a_1} = 3 - 1 = 2$         โฆ..(1)

${a_3} - {a_2} = 9 - 3 = 6$           โฆ..(2)

${a_4} - {a_3} = 27 - 9 = 18$           โฆ..(3)

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

x. $a,2a,3a,4a, \ldots$

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

${a_2} - {a_1} = 2a - a = a$              โฆ..(1)

${a_3} - {a_2} = 3a - 2a = a$              โฆ..(2)

${a_4} - {a_3} = 4a - 3a = a$              โฆ..(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 $a$ and common difference $a$.

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

Substituting, $a = a,d = a$in (4)

we get, ${a_n} = a + \left( {n - 1} \right)d$....(5)

${a_5} = a + (5 - 1)a = 5a$

${a_6} = a + (6 - 1)a = 6a$

${a_7} = a + (7 - 1)a = 7a$

xi. $a,{a^2},{a^3},{a^4}, \ldots$

Ans: For the given series, let us check the difference between all consecutive terms and find if they are equal or not.${a_2} - {a_1} = {a^2} - a = a\left( {a - 1} \right)$              โฆ..(1)

${a_3} - {a_2} = {a^3} - {a^2} = {a^2}\left( {a - 1} \right)$         โฆ..(2)

${a_4} - {a_3} = {a^4} - {a^3} = {a^3}\left( {a - 1} \right)$  โฆโฆ..(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.

xii. $\sqrt 2 ,\sqrt 8 ,\sqrt {18} ,\sqrt {32} , \ldots$

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

${a_2} - {a_1} = \sqrt 8 - \sqrt 2 = 2\sqrt 2 - \sqrt 2 = \sqrt 2$     โฆโฆ(1)

${a_3} - {a_2} = \sqrt {18} - \sqrt 8 = 3\sqrt 2 - 2\sqrt 2 = \sqrt 2$   โฆโฆโฆ(2)

${a_4} - {a_3} = \sqrt {32} - \sqrt {18} = 4\sqrt 2 - 3\sqrt 2 = \sqrt 2$.......(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 $a$ and common difference is given by${a_n} = a + (n - 1)d$      โฆ..(4)

Substituting, a = $\sqrt 2$d = $\sqrt 2$ in (4)

we get,

${a_5} = \sqrt 2 + \left( {5 - 1} \right)\sqrt 2 = \sqrt 2 + 4\sqrt 2 = 5\sqrt 2 = \sqrt {50}$

${a_6} = \sqrt 2 + \left( {6 - 1} \right)\sqrt 2 = \sqrt 2 + 5\sqrt 2 = 6\sqrt 2 = \sqrt {72}$

${a_7} = \sqrt 2 + \left( {7 - 1} \right)\sqrt 2 = \sqrt 2 + 6\sqrt 2 = 7\sqrt 2 = \sqrt {98}$

xiii. $\sqrt 3 ,\sqrt 6 ,\sqrt 9 ,\sqrt {12} , \ldots$

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

${a_2} - {a_1} = \sqrt 6 - \sqrt 3 = \sqrt 3 \times \sqrt 2 - \sqrt 3 = \sqrt 3 \left( {\sqrt 2 - 1} \right)$     โฆโฆ(1)

${a_3} - {a_2} = \sqrt 9 - \sqrt 6 = 3 - \sqrt 6 = \sqrt 3 \left( {\sqrt 3 - \sqrt 2 } \right)$   โฆโฆโฆ(2)

${a_4} - {a_3} = \sqrt {12} - \sqrt 9 = 2\sqrt 3 - 3 = \sqrt 3 \left( {2 - \sqrt 3 } \right)$โฆโฆโฆ.(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.

xiv.  ${1^2},{3^2},{5^2},{7^2}, \ldots$

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

${a_2} - {a_1} = {3^2} - {1^2} = 8$     โฆโฆ(1)

${a_3} - {a_2} = {5^2} - {3^2} = 16$  โฆโฆโฆ(2)

${a_4} - {a_3} = {7^2} - {5^2} = 24$โฆโฆโฆ.(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.

xv.${1^2},{5^2},{7^2},73,...$

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

${a_2} - {a_1} = {5^2} - {1^2} = 24$ โฆโฆ(1)

${a_3} - {a_2} = {7^2} - {5^2} = 24$โฆโฆโฆ(2)

${a_4} - {a_3} = 73 - {7^2} = 24$  โฆโฆโฆ.(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 a and common difference is given by${a_n} = a + (n - 1)d$      โฆ..(4)

Substituting a = ${1^2}$, d=24 in (4)

we get,

${a_5} = {1^2} + \left( {5 - 1} \right)24 = 1 + \left( 4 \right)24 = 97$

${a_6} = {1^2} + \left( {6 - 1} \right)24 = 1 + \left( 5 \right)24 = 121$

${a_7} = {1^2} + \left( {7 - 1} \right)24 = 1 + \left( 6 \right)24 = 145$

## Conclusion

Exercise 5.1 NCERT Solutions for Class 10 Maths Chapter on Arithmetic Progressions provided by Vedantu offers a comprehensive understanding of this fundamental concept. Students can gain a solid understanding of arithmetic progressions by concentrating on important concepts such as calculating the sum of terms, nth term, and common difference. Arithmetic progressions are fundamental to more advanced mathematical ideas, hence it's important to understand them. Practicing a variety of problems from NCERT Solutions and previous years papers can enhance preparation and confidence for exams.

## Class 10 Maths Chapter 5: Exercises Breakdown

 Exercises Number of Questions Exercise 5.2 20 Questions & Solutions (10 Short Answers, 10 Long Answers) Exercise 5.3 20 Questions & Solutions (7 Short Answers, 13 Long Answers) Exercise 5.4 5 Questions & Solutions (5 Long Answers)

## Chapter-Specific NCERT Solutions for Class 10 Maths

Given below are the chapter-wise NCERT Solutions for Class 10 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

## FAQs on NCERT Solutions For Class 10 Maths Chapter 5 Exercise 5.1 - Arithmetic Progressions

1. What is an arithmetic progression in Class 10 Maths Ex 5.1?

In Class 10th Maths Chapter 5 Exercise 5.1 Arithmetic Progression is a series of numbers with a constant difference between terms is called an arithmetic progression (AP). The common difference (d) refers to this constant difference where $(d = a_2 - a_1)$.

2. What is the AP formula for Class 10?

By using this formula we can find nth term in the Ex 5.1 Class 10, Arithmetic Progression

$a_n = a + (n - 1)d$

The sum of the first n terms $S_n$ in an AP is calculated using the formula

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

3. How many examples and questions are there in Exercise 5.1 of Chapter 5 of Class 10 Maths?

There are a total of 4 questions in Exercise 5.1 of Class 10 Maths, and all the questions have sub-parts. You can refer to Vedantu NCERT solutions to get a good idea of the different ways to solve the problems and practice them for grasping the concept properly. These solutions, curated by the subject matter experts at Vedantu, provide the best in class answers to all questions and promise to clear out all your concepts effectively and efficiently. You will be able to understand the concepts well if you download these free of cost solutions PDF.

4. Which is the most favourite question of Exercise 5.1 of 10 Maths?

Chapter 5 of Class 10 Maths talks about Arithmetic Progressions, which is a really interesting topic. All the questions in this chapter can be tricky. Hence, students would have to think logically while solving each question, which is why this might very well be one of the favourite chapters for most students. To know your favourite question from Exercise 5.1 of Class 10 Maths, you must study your Chapter thoroughly with the help of the NCERT solutions by Vedantu. This way you will be clear with your concepts and enjoy your study time.

5. What are the basics of Chapter 5 of Class 10 Maths?

Chapter 5 of Class 10 Maths discusses the concept of Arithmetic Progressions (AP). An AP is actually a sequential order of numbers in which the difference between any two successive terms is a constant. Arithmetic Progressions is undoubtedly a tricky topic and requires a lot of practice along with proper guidance. This means that you can excel in your exams only if you have an authentic source of information such as NCERT Solutions for Class 10 Maths Chapter 5 provided by Vedantu.

6. Do I need to practice all the questions provided in Chapter 5 of Class 10 Maths NCERT Solutions?

All the questions are extremely important in Chapter 5 Class 10 Maths NCERT solutions PDF provided by Vedantu. Therefore, it is crucial to practice every single one of them thoroughly in order to understand the concepts completely. The NCERT solutions by Vedantu are guaranteed to provide you with an edge over your peers and help you to practice well to ace your Class 10 Maths Board exams efficiently. These answers are tailored to perfection to suit the needs of students from the exam point of view.

7. What is the best Solution book for Chapter 5 of NCERT Class 10th Maths?

NCERT solutions available on Vedantu app and website are the best source to get the solutions for Chapter 5 of Class 10 Maths. These solutions, curated by the subject matter experts, are accurate and effective for the exams and allow the students to clarify their doubts for an even better result. These solutions are the best in the market and are free to download, which means you can download and study from them according to your comfort and convenience. This will help you to focus on all your subjects while securing a sure shot high score in Maths.