NCERT Solutions Class 10 Mathematics Chapter 5 Arithmetic Progressions

These NCERT Solutions Class 10 Mathematics Chapter 5 Arithmetic Progressions provide comprehensive, step-by-step guidance tailored for CBSE board exam preparation. In the Class 10 syllabus, Chapter 5 introduces students to the algebraic concept of numerical sequences that increase or decrease by a constant amount. Arithmetic Progressions (AP) forms a core pillar of the Algebra unit, which collectively carries 20 marks in the CBSE Class 10 Mathematics annual board examination. Chapter 5 specifically accounts for 4 to 6 marks across multiple-choice questions, short-answer questions, and 4-mark real-world case studies. This comprehensive guide covers all three primary exercises (Exercise 5.1, Exercise 5.2, and Exercise 5.3), offering step-by-step mathematical reasoning, complete derivations, key formulas, common pitfalls, and practical shortcuts.

Chapter Overview

An Arithmetic Progression (AP) is a sequence of numbers in which each term after the first is obtained by adding a fixed number, known as the common difference (denoted by d), to the preceding term. Arithmetic Progressions model various real-life phenomena such as salary increments, installment repayments, ladder rung spacing, and simple interest growth.

Key definitions and concepts covered in Chapter 5 include:

• First Term (a): The initial number of the arithmetic sequence, denoted as a or a₁.
• Common Difference (d): The constant difference between any term and its preceding term: d = a_k+1 − a_k. The common difference can be positive, negative, or zero.
• General Form of an AP: a, a + d, a + 2d, a + 3d, …, a + (n − 1)d.
• Finite vs. Infinite AP: An AP containing a finite number of terms has a distinct last term (denoted by l), whereas an infinite AP continues indefinitely without a terminal value.

The NCERT Class 10 Mathematics Chapter 5 curriculum is systematically divided into three main exercises:

• Exercise 5.1: Fundamentals of AP — identifying arithmetic sequences, calculating the first term and common difference, and formulating terms from given conditions.
• Exercise 5.2: The n^th Term of an AP — applying the general term formula a_n = a + (n − 1)d to find missing terms, term indices, and terms from the end.
• Exercise 5.3: Sum of the First n Terms — deriving and applying S_n = (n / 2)[2a + (n − 1)d] and solving practical word problems.

Exercise Solutions

Below are detailed, step-by-step solutions for representative and high-frequency questions across all exercises of NCERT Class 10 Mathematics Chapter 5.

Exercise 5.1 — Understanding Arithmetic Progressions

Question 1 (NCERT Page 99): In which of the following situations does the list of numbers involved make an arithmetic progression, and why?

Situation: The taxi fare after each km when the fare is ₹15 for the first km and ₹8 for each additional km.

• Given: Fare for the 1^st km (a₁) = ₹15; Fare for each additional km = ₹8.
• To Find: Whether the sequence of total fares forms an AP.
• Step-by-Step Calculation:
  1. Fare for 1 km: a₁ = 15
  2. Fare for 2 km: a₂ = 15 + 8 = ₹23
  3. Fare for 3 km: a₃ = 23 + 8 = ₹31
  4. Fare for 4 km: a₄ = 31 + 8 = ₹39
  5. Check difference: a₂ − a₁ = 23 − 15 = 8; a₃ − a₂ = 31 − 23 = 8; a₄ − a₃ = 39 − 31 = 8.
• Conclusion: Since the difference (a_k+1 − a_k) is constant (d = 8), this situation forms an AP.
• Final Answer: Yes, the sequence 15, 23, 31, 39, … forms an AP with a = 15 and d = 8.

Question 4 (NCERT Page 100): Which of the following are APs? If they form an AP, find the common difference d and write three more terms: √2, √8, √18, √32, …

• Given: Sequence √2, √8, √18, √32, …
• Step-by-Step Simplification:
  1. a₁ = √2
  2. a₂ = √8 = √(4 × 2) = 2√2
  3. a₃ = √18 = √(9 × 2) = 3√2
  4. a₄ = √32 = √(16 × 2) = 4√2
• Calculate Differences:
  1. a₂ − a₁ = 2√2 − √2 = √2
  2. a₃ − a₂ = 3√2 − 2√2 = √2
  3. a₄ − a₃ = 4√2 − 3√2 = √2
• Next Three Terms:
  1. a₅ = 4√2 + √2 = 5√2 = √50
  2. a₆ = 5√2 + √2 = 6√2 = √72
  3. a₇ = 6√2 + √2 = 7√2 = √98
• Final Answer: It is an AP with common difference d = √2. The next three terms are √50, √72, and √98.

Exercise 5.2 — The nth Term of an AP

Question 7 (NCERT Page 106): Find the 31^st term of an AP whose 11^th term is 38 and the 16^th term is 73.

• Given: 11^th term (a₁₁) = 38; 16^th term (a₁₆) = 73.
• To Find: 31^st term (a₃₁).
• Formula: a_n = a + (n − 1)d
• Step-by-Step Calculation:
  1. Set up Equation (1): a₁₁ = a + (11 − 1)d ⇒ a + 10d = 38 … (1)
  2. Set up Equation (2): a₁₆ = a + (16 − 1)d ⇒ a + 15d = 73 … (2)
  3. Subtract Equation (1) from Equation (2): (a + 15d) − (a + 10d) = 73 − 38 ⇒ 5d = 35 ⇒ d = 7
  4. Substitute d = 7 into Equation (1): a + 10(7) = 38 ⇒ a + 70 = 38 ⇒ a = −32
  5. Calculate a₃₁: a₃₁ = a + (31 − 1)d = −32 + 30(7) = −32 + 210 = 178
• Final Answer: The 31^st term of the AP is 178.

Question 13 (NCERT Page 106): How many three-digit numbers are divisible by 7?

• Given: Three-digit numbers range from 100 to 999.
• To Find: Number of terms (n) divisible by 7.
• Step-by-Step Calculation:
  1. Smallest 3-digit number divisible by 7: 100 ÷ 7 = 14 with remainder 2. Smallest number = 100 + (7 − 2) = 105. Thus, a = 105.
  2. Largest 3-digit number divisible by 7: 999 ÷ 7 = 142 with remainder 5. Largest number = 999 − 5 = 994. Thus, a_n = l = 994.
  3. Common difference: d = 7.
  4. Sequence: 105, 112, 119, …, 994.
  5. Formula substitution: a_n = a + (n − 1)d ⇒ 994 = 105 + (n − 1)7
  6. 994 − 105 = 7(n − 1) ⇒ 889 = 7(n − 1)
  7. n − 1 = 889 / 7 = 127 ⇒ n = 128
• Final Answer: There are 128 three-digit numbers divisible by 7.

Question 17 (NCERT Page 107): Find the 20^th term from the last term of the AP: 3, 8, 13, …, 253.

• Given: AP = 3, 8, 13, …, 253; Last term (l) = 253; Common difference (d) = 8 − 3 = 5.
• Formula: n^th term from the end = l − (n − 1)d
• Step-by-Step Calculation:
  1. Here, l = 253, n = 20, and d = 5.
  2. 20^th term from the end = 253 − (20 − 1) × 5
  3. = 253 − (19 × 5)
  4. = 253 − 95 = 158
• Final Answer: The 20^th term from the last term is 158.

Exercise 5.3 — Sum of First n Terms of an AP

Question 9 (NCERT Page 113): If the sum of the first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of the first n terms.

• Given: S₇ = 49 and S₁₇ = 289.
• To Find: S_n in terms of n.
• Formula: S_n = (n / 2)[2a + (n − 1)d]
• Step-by-Step Calculation:
  1. For n = 7: S₇ = (7 / 2)[2a + 6d] = 49 ⇒ (7 / 2) × 2[a + 3d] = 49 ⇒ a + 3d = 7 … (1)
  2. For n = 17: S₁₇ = (17 / 2)[2a + 16d] = 289 ⇒ (17 / 2) × 2[a + 8d] = 289 ⇒ a + 8d = 17 … (2)
  3. Subtract (1) from (2): (a + 8d) − (a + 3d) = 17 − 7 ⇒ 5d = 10 ⇒ d = 2
  4. Substitute d = 2 into (1): a + 3(2) = 7 ⇒ a + 6 = 7 ⇒ a = 1
  5. Find S_n: S_n = (n / 2)[2(1) + (n − 1)(2)] = (n / 2)[2 + 2n − 2] = (n / 2)[2n] = n²
• Final Answer: The sum of the first n terms is S_n = n².

Question 11 (NCERT Page 113): If the sum of the first n terms of an AP is given by S_n = 4n − n², what is the first term (S₁)? What is the sum of the first two terms? What is the second term? Similarly, find the 3^rd, 10^th, and n^th terms.

• Given: S_n = 4n − n²
• Key Identity: a₁ = S₁ and a_n = S_n − S_n−1 for n ≥ 2
• Step-by-Step Calculation:
  1. First term: a₁ = S₁ = 4(1) − (1)² = 4 − 1 = 3
  2. Sum of first two terms: S₂ = 4(2) − (2)² = 8 − 4 = 4
  3. Second term: a₂ = S₂ − S₁ = 4 − 3 = 1
  4. Common difference: d = a₂ − a₁ = 1 − 3 = −2
  5. Third term: a₃ = a + 2d = 3 + 2(−2) = 3 − 4 = −1
  6. Tenth term: a₁₀ = a + 9d = 3 + 9(−2) = 3 − 18 = −15
  7. General n^th term: a_n = S_n − S_n−1 = (4n − n²) − [4(n − 1) − (n − 1)²]
     = 4n − n² − [4n − 4 − (n² − 2n + 1)] = 4n − n² − [6n − n² − 5] = 5 − 2n
• Final Answer: First term = 3, Sum of first 2 terms = 4, Second term = 1, 3^rd term = −1, 10^th term = −15, and n^th term = 5 − 2n.

Important Formulas and Theorems

Mastering these mathematical relationships is essential for solving any NCERT Mathematics Class 10 solutions problem quickly and accurately during exams:

Derivation of Sum of First n Terms (Gauss's Method)

Let an AP have first term a and common difference d. Writing the sum S_n in forward and reverse order:

1. S_n = a + (a + d) + (a + 2d) + … + [a + (n − 1)d] … (Equation 1)
2. S_n = [a + (n − 1)d] + [a + (n − 2)d] + … + (a + d) + a … (Equation 2)
3. Adding Equation (1) and Equation (2) term-by-term vertically yields n identical terms of [2a + (n − 1)d]:
4. 2S_n = n × [2a + (n − 1)d]
5. Dividing by 2 gives: S_n = (n / 2)[2a + (n − 1)d]

Choosing Convenient Terms in AP for Algebraic Problems

When solving word problems involving symmetric relations, select symmetric variables to simplify calculations:

• 3 terms in AP: Let the terms be (a − d), a, (a + d) ⇒ Sum = 3a (eliminates d immediately).
• 4 terms in AP: Let the terms be (a − 3d), (a − d), (a + d), (a + 3d) ⇒ Common difference is 2d; Sum = 4a.
• 5 terms in AP: Let the terms be (a − 2d), (a − d), a, (a + d), (a + 2d) ⇒ Sum = 5a.

Common Mistakes and Tips

Avoid these frequent calculation and conceptual errors identified by CBSE evaluators:

• Confusing the Term Number (n) with the Value of the Term (a_n): The position n must always be a positive integer (natural number ∈ ℕ). If your algebraic calculation gives n as a negative number or fraction (e.g., n = −5 or n = 14.3), that term does not exist in the AP. In contrast, a_n, a, and d can be fractions, decimals, or negative numbers.
• Sign Errors with Negative Common Difference: When d is negative, always write substitutions inside parentheses: a + (n − 1)(−3). Omitting brackets leads to writing a + n − 1 − 3, which produces incorrect results.
• Misunderstanding the Double Answer for n in Sum Problems: In problems where S_n yields two positive integer values of n (e.g., n = 4 and n = 13 for an AP with negative d), both answers can be mathematically valid. This occurs because the sum of terms from the 5^th to the 13^th position equals zero. Always write an explanation for both values to secure full marks.
• Forgetting to Reverse Difference When Finding Terms from the End: When calculating from the last term backward, the common difference becomes −d. Alternatively, directly use the standard formula a_n(end) = l − (n − 1)d.

Board Exam Relevance

Arithmetic Progressions is among the most reliable, scoring chapters in the CBSE Class 10 Board Exam. Questions are framed according to the following typical mark distribution:

• Section A (1 Mark - MCQ / Assertion-Reason): Identifying whether a sequence is an AP, finding the common difference when given consecutive algebraic expressions such as (k − 1), (k + 3), (3k − 1), or finding a₁₀ − a₅.
• Section B (2 Marks - Short Answer): Determining which term of an AP is a given number, finding missing box terms in NCERT sequences, or finding the 10^th term from the end.
• Section C (3 Marks - Short Answer): Simultaneous linear equations in a and d (e.g., given two separate terms, find a third term), or calculating sums like the sum of all integers between 100 and 500 divisible by 8.
• Section D / E (4 Marks - Case Study / Long Answer): Real-life application scenarios such as production of TV sets in 3^rd and 7^th years, saving schemes, triangular arrangements of flower beds, or stadium seating rows.

More NCERT Solutions and Practice

To master Arithmetic Progressions and achieve a 100/100 in your Class 10 Board Exams, structured practice of standard NCERT questions, exemplar problems, and previous years' board papers (PYQs) is essential. Explore comprehensive chapter-wise tests, instant AI-evaluated step marking, and customized mock test generation for Class 10 Mathematics at qptool.theorify.in.