NCERT Solutions Class 10 Maths Chapter 5 Exercise 5.1 Arithmetic Progressions

NCERT Solutions for Class 10 Maths Chapter 5 – Arithmetic Progressions Exercise 5.1 are designed to help students with their Class 10 exam preparations. These solutions are prepared by subject experts to explain the fundamental concepts of Arithmetic Progressions (AP), such as finding the nth term, understanding the common difference, and solving problems related to the sum of the first n terms.

Exercise 5.1 in Arithmetic Progressions introduces students to the basic concepts of AP, where each term is obtained by adding a constant common difference to the previous term. The exercise covers simple problems that help students build a solid foundation and prepare for more complex problems in subsequent exercises.

NCERT Solutions Class 10 Maths Chapter 5 Exercise 5.1 Arithmetic Progressions

Arithmetic Progression

Medium

Q.

$\begin{array}{l}{Write first four terms of the AP, when the first term}\\ {a and the common difference d are given as follows:}\\ {(i)   }\mathrm{a}=10,{  }\mathrm{d}=10{ }\\ {(ii)  }\mathrm{a}=-2,{  }\mathrm{d}=0\\ {(iii) }\mathrm{a}=4,{    }\mathrm{d}=-3{ }\\ {(iv) }\mathrm{a}=-1,{  }\mathrm{d}=\frac{1}{2}\\ {(v)  }\mathrm{a}=-1.25,{  }\mathrm{d}=-0.25\end{array}$

View Solution

Arithmetic Progression

Medium

Q.

$\begin{array}{l}{For the following APs, write the first term and the}\\ {common difference:}\\ {(i) }3,{\hspace{0.17em}\hspace{0.17em}}1,{\hspace{0.17em}}-1,{\hspace{0.17em}}-3,{ }.{ }.{ }.{             (ii)}-{5,\hspace{0.17em}\hspace{0.17em}}-{1, 3, 7, . . .}\\ {(iii) }\frac{{1}}{3}{,\hspace{0.17em}\hspace{0.17em}}\frac{{5}}{3}{, }\frac{{9}}{3}{,\hspace{0.17em}\hspace{0.17em}}\frac{{13}}{3}{,. . .         (iv) 0.6, 1.7, 2.8, 3.9, . . .}\end{array}$

View Solution

Arithmetic Progression

Medium

Q.

$\begin{array}{l}\text{Which of the following are APs? If they form an AP,}\\ \text{find the common difference d and write three more}\\ \text{terms.}\\ \text{(i) 2, 4, 8, 16, . . .}\\ \text{(ii) 2, }\frac{\text{5}}{2}\text{, 3, }\frac{\text{7}}{2}\text{, . . .}\\ \text{(iii) }-\text{1.2, }-\text{3.2, }-\text{5.2, }-\text{7.2, . . .}\\ \text{(iv) }-\text{10, }-\text{6,}-\text{2, 2, . . .}\\ \text{(v) 3, }3+\sqrt{2},3+2\sqrt{2},3+3\sqrt{2},...\\ \text{(vi) }0.2,0.22,0.222,0.2222,...\\ \text{(vii) 0, }-\text{4, }-\text{8, }-\text{12, . . .}\\ \text{(viii) }-\frac{\text{1}}{2}\text{, }-\frac{\text{1}}{2}\text{, }-\frac{\text{1}}{2}\text{, }-\frac{\text{1}}{2}\text{, . . .}\\ \text{(ix) 1, 3, 9, 27, }...\\ \text{(x) a, 2a, 3a, 4a, }...\\ {\text{(xi) a, a}}^2,{\text{ a}}^3,{\text{ a}}^4,...\\ \text{(xii) }\sqrt{2},\sqrt{8},\sqrt{18},\sqrt{32},...\\ \text{(xiii) }\sqrt{3},\sqrt{6},\sqrt{9},\sqrt{12},...\\ {\text{(xiv) 1}}^2{\text{, 3}}^2,{\text{ 5}}^2,{\text{ 7}}^2,...\\ {\text{(xv) 1}}^2{\text{, 5}}^2,{\text{ 7}}^2,\text{ 73},...\end{array}$

View Solution

Arithmetic Progression

Medium

Q.

$\begin{array}{l}\text{In which of the following situations, does the list of}\\ \text{numbers involved make an arithmetic progression,}\\ \text{and why?}\\ \text{(i) The taxi fare after each km when the fare is }₹\text{15}\\ \text{ for the first km and }₹\text{8 for each additional km.}\\ \text{(ii) The amount of air present in a cylinder when a }\\ \text{ vacuum pump removes }\frac{1}{4}\text{ of the air remaining}\\ \text{ in the cylinder at a time.}\\ \text{(iii) The cost of digging a well after every metre of }\\ \text{ digging, when it costs }₹\text{150 for the first metre}\\ \text{ and rises by }₹\text{50 for each subsequent metre.}\\ \text{(iv) The amount of money in the account every year,}\\ \text{ when }₹\text{10000 is deposited at compound interest}\\ \text{ at 8 \% per annum.}\end{array}$

View Solution

These solutions are aligned with the latest CBSE syllabus and provide step-by-step guidance, ensuring that students are well-prepared for their exams.

Here is the text from NCERT Solutions Class 10 Maths Chapter 5, Exercise 5.1:

Q1. In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?

i) Taxi fare after each km:

• First km fare = Rs. 15

• Fare for each additional km = Rs. 8

  • Solution:

    • 1st km = Rs. 15

    • 2nd km = 15 + 8 = Rs. 23

    • 3rd km = 15 + 8 + 8 = Rs. 31

    • The difference between the fares is constant (Rs. 8), so this forms an Arithmetic Progression (AP).

ii) The amount of air present in a cylinder:

• A vacuum pump removes 1/4 of the air each time.

  • Solution:

    • This does not form an AP. The removal process doesn't follow a fixed difference; it’s a fraction of the remaining air.

iii) The cost of digging a well:

• First meter cost = Rs. 150, increases by Rs. 50 for each subsequent meter.

  • Solution:

    • 1st meter = Rs. 150

    • 2nd meter = Rs. 150 + 50 = Rs. 200

    • 3rd meter = Rs. 150 + 50 + 50 = Rs. 250

    • The difference (Rs. 50) is constant, so this forms an AP.

iv) The amount of money in a bank account with compound interest:

• Principal = Rs. 10,000, interest = 8% per annum.

  • Solution:

    • The money increases based on compound interest, not a fixed amount, so this does not form an AP.

Q2. Write the first four terms of the following APs:

i) a = 10, d = 10:

• First term = 10

• Second term = 10 + 10 = 20

• Third term = 10 + 10 + 10 = 30

• Fourth term = 10 + 10 + 10 + 10 = 40

• Answer: The first four terms are 10, 20, 30, 40.

ii) a = -2, d = 0:

• First term = -2

• Second term = -2

• Third term = -2

• Fourth term = -2

• Answer: The first four terms are -2, -2, -2, -2.

iii) a = 4, d = -3:

• First term = 4

• Second term = 4 - 3 = 1

• Third term = 4 - 3 - 3 = -2

• Fourth term = 4 - 3 - 3 - 3 = -5

• Answer: The first four terms are 4, 1, -2, -5.

iv) a = 1, d = -2:

• First term = 1

• Second term = 1 - 2 = -1

• Third term = 1 - 2 - 2 = -3

• Fourth term = 1 - 2 - 2 - 2 = -5

• Answer: The first four terms are 1, -1, -3, -5.

v) a = 1.25, d = 0.25:

• First term = 1.25

• Second term = 1.25 + 0.25 = 1.5

• Third term = 1.25 + 0.25 + 0.25 = 1.75

• Fourth term = 1.25 + 0.25 + 0.25 + 0.25 = 2

• Answer: The first four terms are 1.25, 1.5, 1.75, 2.

Q3. For the following APs, write the first term and the common difference:

i) 3, 1, -1, -3

• First term: 3

• Common difference: -2

• Answer: The AP is 3, 1, -1, -3 with common difference -2.

ii) 5, 1, 3, 7

• First term: 5

• Common difference: 4

• Answer: The AP is 5, 1, 3, 7 with common difference 4.

iii) 1, 5, 9, 13

• First term: 1

• Common difference: 4

• Answer: The AP is 1, 5, 9, 13 with common difference 4.

iv) 0.6, 1.7, 2.8, 3.9

• First term: 0.6

• Common difference: 1.1

• Answer: The AP is 0.6, 1.7, 2.8, 3.9 with common difference 1.1.

FAQs: Class 10 Maths Chapter 5 – Arithmetic Progressions Exercise 5.1

Q1. What is an Arithmetic Progression (AP)?
Answer:
An Arithmetic Progression (AP) is a sequence of numbers in which the difference between any two consecutive terms is constant. This constant difference is called the common difference (d).

For example:

$3, 7, 11, 15, \dots$

3,7,11,15,… is an AP with a common difference of 4.

Q2. What is the formula for the nth term of an AP?
Answer:
The nth term of an AP is given by the formula:

$a_n = a_1 + (n - 1) \times d$

an​=a1​+(n−1)×d
where:

• $a_n$an​ is the nth term,

• $a_1$a1​ is the first term,

• $n$n is the term number,

• $d$d is the common difference.

Q3. How do you find the sum of the first n terms of an AP?
Answer:
The sum of the first n terms of an AP is given by the formula:

$S_n = \frac{n}{2} \times \left[ 2a_1 + (n - 1) \times d \right]$

Sn​=2n​×[2a1​+(n−1)×d]
or

$S_n = \frac{n}{2} \times \left[ a_1 + a_n \right]$

Sn​=2n​×[a1​+an​]
where:

• $S_n$Sn​ is the sum of the first n terms,

• $a_1$a1​ is the first term,

• $a_n$an​ is the nth term,

• $d$d is the common difference,

• $n$n is the number of terms.

Q4. How do NCERT Solutions help with exam preparation?
Answer:
These solutions provide clear, step-by-step explanations for solving problems related to arithmetic progressions. Practicing these solutions will help students grasp the key concepts of AP, including finding the nth term, the common difference, and the sum of the first n terms, ensuring better problem-solving skills and strong preparation for Class 10 exams.

Q5. How do I find the common difference in an AP?
Answer:
The common difference d is found by subtracting the first term from the second term (or any two consecutive terms):

$d = a_2 - a_1$

d=a2​−a1​
For example, in the AP

$3, 7, 11, 15$

3,7,11,15, the common difference is

$7 - 3 = 4$

7−3=4.