CBSE Class 10 Mathematics: NCERT Solutions Class 10 Maths Chapter 5 Arithmetic Progressions (2025–2026 Guide)

Mastering CBSE Class 10 Mathematics is essential for securing high scores in your 2025–2026 board examinations. Chapter 5, Arithmetic Progressions (AP), is one of the most scoring and concept-heavy units in the syllabus. This detailed guide to NCERT Solutions Class 10 Maths Chapter 5 Arithmetic Progressions breaks down all fundamental definitions, standard derivation formulas, real-world applications, and high-yield questions to ensure complete conceptual clarity for your board exam 10 preparation.

Key Concepts and Formulas in Arithmetic Progressions

An Arithmetic Progression (AP) is a sequence of numbers in which each term after the first is obtained by adding a fixed number to the preceding term. This fixed number is known as the common difference.

1. Standard Notation and General Form

If the first term of an AP is denoted by a and the common difference by d, the general form of an AP is expressed as:

a, a + d, a + 2d, a + 3d, ..., a + (n - 1)d

• First term: a₁ = a
• Common difference (d): d = a_k+1 - a_k for any positive integer k. The value of d can be positive, negative, or zero.
• Finite vs Infinite AP: An AP with a limited number of terms has a distinct last term (denoted as l), whereas an infinite AP continues indefinitely without a terminal term.

2. The n^th Term (General Term) of an AP

To determine any term in an arithmetic sequence without listing all preceding terms, use the general term formula:

a_n = a + (n - 1)d

Where:

• a_n = the n^th term
• a = the first term
• n = total number of terms (n ∈ ℕ, positive integer)
• d = common difference

Special Case (n^th Term from the End): If an AP has a last term l and a common difference d, the n^th term from the end is given by:

a_n(end) = l - (n - 1)d

3. Sum of the First n Terms (S_n)

The sum of the first n terms of an AP is calculated using either of the following two standard formulas:

1. When the first term (a) and common difference (d) are known:

S_n = (n / 2) × [2a + (n - 1)d]

2. When the first term (a) and last term (l = a_n) are known:

S_n = (n / 2) × (a + l)

Important Relation: The n^th term can also be extracted from consecutive sum values:

a_n = S_n - S_n-1 (for n > 1, with a₁ = S₁)

4. Three Consecutive Terms in an AP

When solving word problems involving three unknown terms in an AP, setting them up symmetrically simplifies calculations:

• Three terms: (a - d), a, (a + d) — with sum = 3a.
• Four terms: (a - 3d), (a - d), (a + d), (a + 3d) — with common difference 2d.

Important CBSE Questions with Answers (Step-by-Step Solutions)

The following problems represent standard question patterns frequently tested in the CBSE Mathematics Class 10 board examinations.

Question 1 (2 Marks — Short Answer)

Find the 20^th term from the last term of the AP: 3, 8, 13, ..., 253.

Solution:

1. Identify the parameters of the given AP:
   • First term (a) = 3
   • Common difference (d) = 8 - 3 = 5
   • Last term (l) = 253
2. Apply the formula for the n^th term from the end:
   an(end) = l - (n - 1)d
3. Substitute n = 20, l = 253, and d = 5:
   a20(end) = 253 - (20 - 1) × 5
a20(end) = 253 - (19 × 5)
a20(end) = 253 - 95 = 158

Answer: The 20^th term from the last term is 158.

Question 2 (3 Marks — Algebraic AP)

If the 7^th term of an AP is 1/9 and its 9^th term is 1/7, find its 63^rd term.

Solution:

1. Let the first term be a and the common difference be d.
2. Formulate equations using an = a + (n - 1)d:
   Equation (1): a + 6d = 1/9
   Equation (2): a + 8d = 1/7
3. Subtract Equation (1) from Equation (2):
   (a + 8d) - (a + 6d) = 1/7 - 1/9
2d = (9 - 7) / 63 = 2 / 63
d = 1 / 63
4. Substitute d = 1/63 into Equation (1):
   a + 6 × (1 / 63) = 1 / 9
a + 2 / 21 = 1 / 9
a = 1/9 - 2/21 = (7 - 6) / 63 = 1 / 63
5. Calculate the 63^rd term:
   a63 = a + (63 - 1)d = a + 62d
a63 = (1 / 63) + 62 × (1 / 63) = (1 + 62) / 63 = 63 / 63 = 1

Answer: The 63^rd term of the AP is 1.

Question 3 (3 Marks — Divisibility and Sum)

How many two-digit positive integers are divisible by 6? Find their total sum.

Solution:

1. List the two-digit numbers divisible by 6:
   The smallest two-digit multiple of 6 is 12, and the largest is 96.
   The sequence is: 12, 18, 24, ..., 96
2. Identify the AP parameters:
   a = 12, d = 6, an = 96
3. Find the number of terms (n):
   a + (n - 1)d = 96
12 + (n - 1)6 = 96
(n - 1)6 = 84 ⇒ n - 1 = 14 ⇒ n = 15
4. Compute the sum using Sn = (n / 2) × (a + l):
   S15 = (15 / 2) × (12 + 96)
S15 = (15 / 2) × 108 = 15 × 54 = 810

Answer: There are 15 two-digit numbers divisible by 6, and their sum is 810.

Question 4 (4 Marks — Quadratic Sum Expression)

The sum of the first n terms of an AP is given by S_n = 3n² + 5n. Find the AP and write its 15^th term.

Solution:

1. Find the first term (a₁):
   a1 = S1 = 3(1)2 + 5(1) = 3 + 5 = 8
2. Find the sum of the first two terms (S₂):
   S2 = 3(2)2 + 5(2) = 3(4) + 10 = 12 + 10 = 22
3. Find the second term (a₂):
   a2 = S2 - S1 = 22 - 8 = 14
4. Determine the common difference (d):
   d = a2 - a1 = 14 - 8 = 6
5. The AP sequence is: 8, 14, 20, 26, ...
6. Calculate the 15^th term (a₁₅):
   a15 = a + 14d = 8 + 14(6) = 8 + 84 = 92

Answer: The AP is 8, 14, 20, 26, ... and its 15^th term is 92.

Question 5 (4 Marks — Case-Based Real-Life Application)

A manufacturer of TV sets produced 600 sets in the third year and 700 sets in the seventh year. Assuming that the production increases uniformly by a fixed number every year, find: (i) the production in the 1^st year, (ii) the production in the 10^th year, and (iii) the total production in first 7 years.

Solution:

1. Since production increases uniformly, yearly production forms an AP with first term a and common difference d.
2. Given conditions:
   a3 = a + 2d = 600 — (Equation 1)
   a7 = a + 6d = 700 — (Equation 2)
3. Subtract Equation (1) from Equation (2):
   4d = 100 ⇒ d = 25
4. Substitute d = 25 in Equation (1):
   a + 2(25) = 600 ⇒ a + 50 = 600 ⇒ a = 550
   (i) Production in the 1^st year = 550 TV sets.
5. Calculate production in the 10^th year (a₁₀):
   a10 = a + 9d = 550 + 9(25) = 550 + 225 = 775
   (ii) Production in the 10^th year = 775 TV sets.
6. Calculate total production in the first 7 years (S₇):
   S7 = (7 / 2) × (a + a7) = (7 / 2) × (550 + 700) = (7 / 2) × 1250 = 7 × 625 = 4375
   (iii) Total production in first 7 years = 4,375 TV sets.

How to Prepare for This Topic in CBSE Class 10 Board Exams

To score a full 100% in questions based on Arithmetic Progressions in your board exam 10, structure your study routine with the following proven strategies:

1. Map Given Information to Variables First: Before applying any formula, write out explicit values for a, d, n, an, and Sn. This prevents variable confusion in multi-step problems.
2. Handle Negative Common Differences Carefully: When an AP is decreasing (e.g., 20, 17, 14, ...), d = -3. Always use parentheses around negative values in formulas to prevent sign errors during multiplication.
3. Validate Quadratic Roots for 'n': When solving for n using the sum formula, you will often obtain a quadratic equation with two roots. Remember that n represents a term index and must be a positive integer (natural number). Reject fractional or negative roots with a clear written reason in your exam copy.
4. Double-Check Symmetrical Choices for Unknown Terms: When the sum of three terms is given, using (a - d), a, (a + d) eliminates d instantly, leaving 3a = Sum. This saves crucial time during timed board examinations.
5. Master Real-World Word Problems: Section E of the CBSE Class 10 Mathematics paper features Case-Based Questions often drawn from Arithmetic Progressions (savings schemes, row arrangements, sports drills). Practice extracting AP parameters from descriptive scenarios.

Where to Practice More

Consistent practice with real board-level problems is the key to mastering CBSE Mathematics. Generate customized, chapter-wise mock tests, access verified marking schemes, and practice authentic previous years' questions (PYQs) for Class 10 Arithmetic Progressions at qptool.theorify.in.