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

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 $8$. 

2. 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. 

3. 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 $50$. 

4. 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 ${{a}_{4}}=40$.

2. \[a=-2,d=0\]

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 ${{a}_{4}}=-2$.

3. \[a=4,d=-3\] 

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 ${{a}_{4}}=-5$.

4. \[a=-1\text{,}d=1/2\] 

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 ${{a}_{4}}=\dfrac{1}{2}$.

5. \[a=-1.25,d=-0.25\] 

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 $=$ $1-3=-2$.  

2. \[-5,-1,3,7,...\] 

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 $=$ $-1-\left( -5 \right)=4$.  

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

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

4. \[0.6,1.7,2.8,3.9,...\] 

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.

2. \[\text{2,}\dfrac{5}{2}\text{,3,}\dfrac{7}{2}\text{,}...\] 

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 (4) 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 ${{a}_{7}}=5$.

3. \[1.2,3.2,5.2,7.2...\] 

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 (4) 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 \[{{a}_{7}}=13.2\].

4. $-10,-6,-2,2,...$ 

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

Therefore, from (5)

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

5. \[3,3+\sqrt{2},3+2\sqrt{2},3+3\sqrt{2},...\] 

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 (4) 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 \[{{a}_{7}}=3+6\sqrt{2}\].

6. \[\text{0}\text{.2,0}\text{.22,0}\text{.222,0}\text{.2222}.....\] 

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.

7. \[0,-4,-8,-12....\] 

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

Therefore, from (5)

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

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

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 (4) 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 \[{{a}_{7}}=-\dfrac{1}{2}\].

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.

${a_2} - {a_1} = 3 - 1 = 2$         …..(1)

${a_3} - {a_2} = 9 - 3 = 6$           …..(2)

${a_4} - {a_3} = 27 - 9 = 18$           …..(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.

${a_2} - {a_1} = 2a - a = a$              …..(1)

${a_3} - {a_2} = 3a - 2a = a$              …..(2)

${a_4} - {a_3} = 4a - 3a = a$              …..(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 ${a_n} = a + \left( {n - 1} \right)d$       …..(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.${a_2} - {a_1} = {a^2} - a = a\left( {a - 1} \right)$              …..(1)

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

${a_4} - {a_3} = {a^4} - {a^3} = {a^3}\left( {a - 1} \right)$  ……..(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 $     ……(1)

${a_3} - {a_2} = \sqrt {18}  - \sqrt 8  = 3\sqrt 2  - 2\sqrt 2  = \sqrt 2 $   ………(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${a_n} = a + (n - 1)d$      …..(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.

${a_2} - {a_1} = \sqrt 6  - \sqrt 3  = \sqrt 3  \times \sqrt 2  - \sqrt 3  = \sqrt 3 \left( {\sqrt 2  - 1} \right)$     ……(1)

${a_3} - {a_2} = \sqrt 9  - \sqrt 6  = 3 - \sqrt 6  = \sqrt 3 \left( {\sqrt 3  - \sqrt 2 } \right)$   ………(2)

${a_4} - {a_3} = \sqrt {12}  - \sqrt 9  = 2\sqrt 3  - 3 = \sqrt 3 \left( {2 - \sqrt 3 } \right)$……….(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.

${a_2} - {a_1} = {3^2} - {1^2} = 8$     ……(1)

${a_3} - {a_2} = {5^2} - {3^2} = 16$  ………(2)

${a_4} - {a_3} = {7^2} - {5^2} = 24$……….(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.

${a_2} - {a_1} = {5^2} - {1^2} = 24$ ……(1)

${a_3} - {a_2} = {7^2} - {5^2} = 24$………(2)

${a_4} - {a_3} = 73 - {7^2} = 24$  ……….(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${a_n} = a + (n - 1)d$      …..(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

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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.