Class 10 Maths Chapter 5 Arithmetic Progressions Exercise 5.1 Solutions | English Medium | SEBA/SCERT Assam 2026–2027

Class 10 Maths Chapter 5 Arithmetic Progressions Exercise 5.1 Solutions | English Medium

Are you searching for the best Class 10 Maths Chapter 5 Arithmetic Progressions Exercise 5.1 Solutions in English Medium? Here you will get complete and step-by-step solutions based on the latest SEBA/SCERT Assam 2026–2027 syllabus.

This chapter is very important for HSLC Examination preparation because students learn sequences, patterns, common difference, and formulas used in Arithmetic Progressions (AP).

Chapter 5 – Arithmetic Progressions

An Arithmetic Progression (AP) is a sequence of numbers in which the difference between consecutive terms remains constant.

Exercise 5.1 – What You Will Learn

In Exercise 5.1, students learn:

How to identify Arithmetic Progressions
Finding common difference
Writing next terms
Checking whether a sequence is AP or not

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Step by step solution for Exercise 5.1  

 

 

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

(i) The taxi fare after each km when the fare is ₹15 for the first km and ₹8 for each additional km.

Solution:

Let \(a_1, a_2, a_3, \dots\) be the taxi fare after 1st, 2nd, 3rd... km respectively.
\(a_1 = 15\)
\(a_2 = 15 + 8 = 23\)
\(a_3 = 23 + 8 = 31\)
\(a_4 = 31 + 8 = 39\)

The list of numbers is: \(15, 23, 31, 39, \dots\)
Since the common difference \(d = 8\) is constant, this forms an Arithmetic Progression (AP).

(ii) The amount of air remaining in a cylinder when a vacuum pump removes \(\frac{1}{4}\) of the air remaining in the cylinder at a time.

Solution:

Let the initial volume of air in the cylinder be = \(V\)
Air remaining after 1st removal (\(a_1\)) = \(V - \frac{1}{4}V = \frac{3}{4}V\)
Air remaining after 2nd removal (\(a_2\)) = \(\frac{3}{4}V - \frac{1}{4}(\frac{3}{4}V) = \frac{3}{4}V - \frac{3}{16}V = \frac{9}{16}V\)
Air remaining after 3rd removal (\(a_3\)) = \(\frac{9}{16}V - \frac{1}{4}(\frac{9}{16}V) = \frac{27}{64}V\)

Here, \(a_2 - a_1 = \frac{9}{16}V - \frac{3}{4}V = -\frac{3}{16}V\)
And \(a_3 - a_2 = \frac{27}{64}V - \frac{9}{16}V = -\frac{9}{64}V\)
Since \(a_2 - a_1 \neq a_3 - a_2\), this does not form an Arithmetic Progression (AP).

(iii) The cost of digging a well after every metre of digging, when it costs ₹150 for the first metre and rises by ₹50 for each subsequent metre.

Solution:

Cost of digging for the 1st metre (\(a_1\)) = ₹150.
Total cost for 2 metres (\(a_2\)) = \(150 + 50 = 200\).
Total cost for 3 metres (\(a_3\)) = \(200 + 50 = 250\).

The list of numbers is: \(150, 200, 250, 300, \dots\)
Since the difference between consecutive terms is constant, \(d = 50\), this forms an Arithmetic Progression (AP).

(iv) The amount of money in the account every year, when ₹10,000 is deposited at compound interest at \(8\%\) per annum.

Solution:

We know that for compound interest, the amount after \(n\) years is \(A = P(1 + \frac{r}{100})^n\).
Here \(P = 10000\) and \(r = 8\).
Amount after 1st year (\(a_1\)) = \(10000(1 + \frac{8}{100})^1\)
Amount after 2nd year (\(a_2\)) = \(10000(1 + \frac{8}{100})^2\)
Amount after 3rd year (\(a_3\)) = \(10000(1 + \frac{8}{100})^3\)

The difference between consecutive terms is not constant because the powers are increasing.
Therefore, this situation does not form an Arithmetic Progression (AP).

 

2. Write the first four terms of the AP, when the first term \(a\) and the common difference \(d\) are given as follows:

Solution: The general form of an AP is: \(a, a+d, a+2d, a+3d, \dots\)

(i) \(a = 10, d = 10\):
First four terms: \(10, 20, 30, 40\).

(ii) \(a = -2, d = 0\):
First four terms: \(-2, -2, -2, -2\).

(iii) \(a = 4, d = -3\):
First four terms: \(4, (4-3), (1-3), (-2-3) \Rightarrow 4, 1, -2, -5\).

(iv) \(a = -1, d = \frac{1}{2}\):
First four terms: \(-1, -\frac{1}{2}, 0, \frac{1}{2}\).

(v) \(a = -1.25, d = -0.25\):
First four terms: \(-1.25, -1.50, -1.75, -2.00\).

3. For the following APs, find the first term and the common difference:

Solution: First term = \(a\), Common difference \(d = a_2 - a_1\)

(i) \(3, 1, -1, -3, \dots\):
First term \(a = 3\), Common difference \(d = 1 - 3 = -2\).

(ii) \(-5, -1, 3, 7, \dots\):
First term \(a = -5\), Common difference \(d = -1 - (-5) = 4\).

(iii) \(\frac{1}{3}, \frac{5}{3}, \frac{9}{3}, \frac{13}{3}, \dots\):
First term \(a = \frac{1}{3}\), Common difference \(d = \frac{5}{3} - \frac{1}{3} = \frac{4}{3}\).

(iv) \(0.6, 1.7, 2.8, 3.9, \dots\):
First term \(a = 0.6\), Common difference \(d = 1.7 - 0.6 = 1.1\).

4. Which of the following are APs? If they form an AP, find the common difference \(d\) and write three more terms.

Method: A sequence is an Arithmetic Progression (AP) if the difference between consecutive terms \(d = a_{n} - a_{n-1}\) is constant.

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(i) \(2, 4, 8, 16, \dots\)
\(a_2 - a_1 = 4 - 2 = 2\)
\(a_3 - a_2 = 8 - 4 = 4\)
Since \(2 \neq 4\), it is not an AP.

(ii) \(2, \frac{5}{2}, 3, \frac{7}{2}, \dots\)
\(a_2 - a_1 = \frac{5}{2} - 2 = \frac{1}{2}\)
\(a_3 - a_2 = 3 - \frac{5}{2} = \frac{1}{2}\)
Here \(d = \frac{1}{2}\). It is an AP.
Next 3 terms: \(4, \frac{9}{2}, 5\)

(iii) \(-1.2, -3.2, -5.2, -7.2, \dots\)
\(a_2 - a_1 = -3.2 - (-1.2) = -2\)
\(a_3 - a_2 = -5.2 - (-3.2) = -2\)
Here \(d = -2\). It is an AP.
Next 3 terms: \(-9.2, -11.2, -13.2\)

(iv) \(-10, -6, -2, 2, \dots\)
\(a_2 - a_1 = -6 - (-10) = 4\)
\(a_3 - a_2 = -2 - (-6) = 4\)
Here \(d = 4\). It is an AP.
Next 3 terms: \(6, 10, 14\)

(v) \(3, 3+\sqrt{2}, 3+2\sqrt{2}, 3+3\sqrt{2}, \dots\)
\(d = (3+\sqrt{2}) - 3 = \sqrt{2}\)
Here \(d = \sqrt{2}\). It is an AP.
Next 3 terms: \(3+4\sqrt{2}, 3+5\sqrt{2}, 3+6\sqrt{2}\)

(vi) \(0.2, 0.22, 0.222, \dots\)
\(a_2 - a_1 = 0.02\), but \(a_3 - a_2 = 0.002\).
Differences are not equal. Not an AP.

(vii) \(0, -4, -8, -12, \dots\)
\(d = -4 - 0 = -4\). It is an AP.
Next 3 terms: \(-16, -20, -24\)

(viii) \(-\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, \dots\)
\(d = 0\). It is an AP.
Next 3 terms: \(-\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}\)

(ix) \(1, 3, 9, 27, \dots\)
\(3-1=2\), while \(9-3=6\). Not an AP.

(x) \(a, 2a, 3a, 4a, \dots\)
\(d = 2a - a = a\). It is an AP.
Next 3 terms: \(5a, 6a, 7a\)

(xi) \(a, a^2, a^3, a^4, \dots\)
\(a^2 - a \neq a^3 - a^2\). Not an AP.

(xii) \(\sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32}, \dots\)
Simplified: \(\sqrt{2}, 2\sqrt{2}, 3\sqrt{2}, 4\sqrt{2}\)
\(d = \sqrt{2}\). It is an AP.
Next 3 terms: \(\sqrt{50}, \sqrt{72}, \sqrt{98}\)

(xiii) \(\sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12}, \dots\)
\(\sqrt{6}-\sqrt{3} \neq \sqrt{9}-\sqrt{6}\). Not an AP.

(xiv) \(1^2, 3^2, 5^2, 7^2, \dots\)
Values: \(1, 9, 25, 49\). Differences are not equal. Not an AP.

(xv) \(1^2, 5^2, 7^2, 73, \dots\)
Values: \(1, 25, 49, 73\)
\(d = 24\). It is an AP.
Next 3 terms: \(97, 121, 145\)

5. Which of the following does not form an Arithmetic Progression (AP)?

(i) \(2,\; 4,\; 6,\; 8,\; 10,\; \ldots\)

(ii) \(2,\; 4,\; 8,\; 16,\; \ldots\)

(iii) \(1,\; -2,\; -5,\; -8,\; \ldots\)

(iv) \(1.8,\; 2.0,\; 2.2,\; 2.4,\; \ldots\)

(A) (i)
(B) (ii)
(C) (iii)
(D) (iv)

Solution:

In an Arithmetic Progression, the difference between consecutive terms is constant.

(i) \(4-2=2,\; 6-4=2\) ⇒ AP

(ii) \(4-2=2,\; 8-4=4\) ⇒ Common difference is not same ⇒ Not an AP

(iii) \(-2-1=-3,\; -5-(-2)=-3\) ⇒ AP

(iv) \(2.0-1.8=0.2,\; 2.2-2.0=0.2\) ⇒ AP

Correct Answer: (B)

6. Consider the statements P to S. The common differences of the Arithmetic Progressions are mixed up. Choose the correct ascending order of the common differences.

P. \(2,\; \frac{5}{2},\; 3,\; \frac{7}{2}\)

Q. \(1.2,\; 3.2,\; 5.2,\; 7.2,\; \ldots\)

R. \(7,\; 10 \frac{1}{2},\; 14,\; 17 \frac{1}{2}\)

S. \(\frac{1}{15},\; \frac{1}{12},\; \frac{1}{10},\; \frac{7}{60}\)

(A) S, Q, R, P
(B) R, P, Q, S
(C) S, P, Q, R
(D) P, Q, S, R

Solution:

P : \(d=\frac{5}{2}-2=\frac{1}{2}\)

Q : \(d=3.2-1.2=2\)

R : \(d=10\frac{1}{2}-7=3\frac{1}{2}\)

S : \(\frac{1}{12}-\frac{1}{15}=\frac{1}{60}\)

In ascending order: \[ \frac{1}{60} < \frac{1}{2} < 2 < 3\frac{1}{2} \]

Therefore, the correct order is: \[ S,\; P,\; Q,\; R \]

Correct Answer: (C)

7. If \(p,\; 2p - 1\) and \(2p + 1\) are consecutive terms of an Arithmetic Progression, then the value of \(p\) is—

(A) \(-3\)
(B) \(-2\)
(C) \(3\)
(D) \(6\)

Solution:

For an AP, \[ (2p-1)-p=(2p+1)-(2p-1) \]

\[ p-1=2 \]

\[ p=3 \]

Correct Answer: (C)

8. Show that \(a-b,\; a,\; a+b\) are consecutive terms of an Arithmetic Progression.

Solution:

Second term − First term \[ = a-(a-b) \]

\[ = a-a+b=b \]

Third term − Second term \[ =(a+b)-a \]

\[ = b \]

Since both differences are equal, \[ a-b,\; a,\; a+b \] are consecutive terms of an Arithmetic Progression.

Why Arithmetic Progression Is Important

Arithmetic Progression is important because it helps students understand:

Number patterns
Sequences
Algebraic thinking
Real-life calculations
Competitive mathematics

This chapter is also useful in higher mathematics and science subjects.

Benefits of These Solutions

✅ Based on latest SEBA/SCERT Assam 2026–2027 syllabus
✅ Easy English explanations
✅ Step-by-step solutions
✅ Board exam focused preparation
✅ Important formulas included
✅ Student-friendly notes

Important Points for HSLC Exam

Students should remember:

Definition of AP
Common Difference Formula
nth Term Formula
Identification of AP
Pattern-based questions

These topics are frequently asked in HSLC board examinations.

Final Words

These Class 10 Maths Chapter 5 Arithmetic Progressions Exercise 5.1 Solutions are designed to help students learn concepts easily and score high marks in the HSLC Examination.

Class 10 Maths Chapter 5 – Arithmetic Progressions Exercise 5.1 FAQs | English Medium

Exercise 5.1 teaches the basic concepts of Arithmetic Progressions (AP), including identifying arithmetic sequences and finding the common difference.

👉 Students learn how numbers increase or decrease in a fixed pattern.

An Arithmetic Progression is a sequence of numbers in which the difference between consecutive terms remains constant.

📘 Example: 2, 5, 8, 11, 14 is an AP because the common difference is 3.

The common difference is the fixed value obtained by subtracting one term from the next term.

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It is represented by the symbol d.

A sequence is an AP if the difference between consecutive terms is always the same.

🎯 Check the difference carefully between every pair of terms.

Arithmetic Progressions is an important chapter for HSLC examinations because formula-based and sequence-related questions are frequently asked.

⭐ Understanding the concepts properly helps students solve sums quickly in exams.

Students often make mistakes while finding the common difference or checking whether the sequence follows a fixed pattern.

🚀 Pro Tip: Always subtract terms carefully to avoid sign mistakes.

Students should practice sequences regularly, understand the formulas properly, and solve previous year HSLC questions.

Digital Pipal Academy provides easy explanations, step-by-step solutions, and updated SEBA/SCERT Assam syllabus-based content for students.

🌟 It helps students improve confidence, speed, and problem-solving skills.

You can find complete step-by-step solutions for Exercise 5.1 on Digital Pipal Academy.