CBSE Class 10 Mathematics Arithmetic Progressions Notes
About This Chapter
Chapter 5 — Arithmetic Progressions is one of the most scoring and conceptually rich chapters in the CBSE Class 10 Mathematics syllabus. It introduces students to the idea of number patterns and sequences that follow a fixed rule — a concept that bridges basic arithmetic and higher algebra.
In this chapter, you will explore what makes a sequence an Arithmetic Progression, how to find any term of the sequence without listing all the terms before it, and how to quickly calculate the sum of a large number of terms using elegant formulas. These ideas have direct real-world applications — from calculating simple interest over multiple years, to predicting the number of seats in a stadium row, to planning equal salary hikes over time.
This chapter typically carries 5 to 8 marks in the CBSE board examination and appears consistently across 1-mark MCQs, 3-mark short answers, and 5-mark long-answer questions. Mastering the two core formulas — the nth term formula and the sum formula — along with their smart applications, is the key to scoring full marks in this section.
What You Will Learn:
• How to identify whether a given sequence is an AP
• Finding the common difference, first term, and nth term of any AP
• Calculating the sum of the first n terms using two powerful formulas
• Applying AP concepts to real-life word problems and board exam questions
• Smart tricks for 3-term and 4-term AP problems to save time in exams
Whether you are preparing for your board exams, revising before tests, or looking for a clear and structured explanation of this chapter, these notes cover everything you need — from concept basics to solved examples, formula summaries, and categorised practice questions.
A detailed and printable PDF version of these notes is attached below for your convenience. Download it, save it, and keep it handy for your revision!
1. Introduction to Arithmetic Progressions
An Arithmetic Progression (AP) is one of the most fundamental sequences in mathematics. It appears in everyday life — from the number of seats in a row of an auditorium, to the pattern of salary increments, to the distance a freely falling body travels. Understanding APs equips students with a powerful mathematical tool used in higher studies, competitive exams, and real-world problem solving.
1.1 What is a Sequence?
A sequence is an ordered list of numbers following a definite rule. Each number in the sequence is called a term. For example: 2, 4, 6, 8, 10, ... is a sequence where each term is obtained by adding 2 to the previous term.
1.2 Definition of Arithmetic Progression
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, denoted by d.
⚑ Key Definition A sequence a1, a2, a3, ..., an is called an Arithmetic Progression if a(n+1) - a(n) = d (constant) for all n = 1, 2, 3, ... |
1.3 Components of an AP
• First Term (a or a1): The starting value of the sequence.
• Common Difference (d): d = a2 - a1 = a3 - a2 = ... (can be positive, negative, or zero).
• Number of Terms (n): Total count of terms in a finite AP.
• Last Term (l or an): The final term in a finite AP.
1.4 Examples of Arithmetic Progressions
Example 1.1 — Identifying APs Sequence: 3, 7, 11, 15, 19, ... Common difference: d = 7 - 3 = 4 Check: 11 - 7 = 4 ✓ 15 - 11 = 4 ✓ This IS an AP with a = 3, d = 4
Sequence: 1, 4, 9, 16, 25, ... Check: 4 - 1 = 3, 9 - 4 = 5 (NOT equal) This is NOT an AP (it is the sequence of perfect squares). |
Type of AP | First Term (a) | Common Difference (d) | Example |
Finite AP | Any real number | Any real number | 2, 5, 8, 11, 14 |
Infinite AP | Any real number | Any real number | 1, 3, 5, 7, 9, ... |
Increasing AP | Any value | d > 0 | 10, 14, 18, 22, ... |
Decreasing AP | Any value | d < 0 | 20, 17, 14, 11, ... |
Constant AP | Any value | d = 0 | 5, 5, 5, 5, 5, ... |
2. General Term (nth Term) of an AP
2.1 Derivation of the General Term Formula
Let the first term of an AP be 'a' and common difference be 'd'. We can write:
• 1st term: a1 = a
• 2nd term: a2 = a + d
• 3rd term: a3 = a + 2d
• 4th term: a4 = a + 3d
• ...
• nth term: an = a + (n - 1)d
Formula — nth Term (General Term) an = a + (n - 1) × d Where: a = first term, d = common difference, n = term number |
2.2 Finding a Specific Term
Example 2.1 — Find the 15th term of the AP: 7, 13, 19, 25, ... Given: a = 7, d = 13 - 7 = 6, n = 15 Using: an = a + (n - 1)d a15 = 7 + (15 - 1) × 6 a15 = 7 + 14 × 6 a15 = 7 + 84 = 91
Answer: The 15th term is 91. |
Example 2.2 — Which term of the AP: 3, 8, 13, 18, ... is 78? Given: a = 3, d = 5, an = 78, find n an = a + (n - 1)d 78 = 3 + (n - 1) × 5 75 = (n - 1) × 5 n - 1 = 15 => n = 16
Answer: 78 is the 16th term of the AP. |
2.3 Checking if a Number Belongs to an AP
To check if a number 'k' is a term of an AP with first term 'a' and common difference 'd', solve for n in: k = a + (n - 1)d. If n is a positive integer, then 'k' is the nth term.
⚑ Important Tip If n comes out as a fraction or negative number, the given value is NOT a term of the AP. |
3. Sum of First n Terms of an AP
3.1 Derivation of Sum Formula
Let Sn denote the sum of the first n terms of an AP with first term 'a' and common difference 'd'. Writing the sum in two ways and adding:
• Sn = a + (a+d) + (a+2d) + ... + (l-d) + l
• Sn = l + (l-d) + (l-2d) + ... + (a+d) + a
• Adding: 2Sn = n × (a + l)
• Therefore: Sn = n/2 × (a + l)
Formula — Sum of First n Terms Sn = n/2 × (2a + (n-1)d) OR equivalently: Sn = n/2 × (a + l) Where: a = first term, d = common difference, l = last term, n = no. of terms |
3.2 Relationship Between an and Sn
⚑ Key Relationship an = Sn - S(n-1) The nth term of an AP equals the sum of n terms minus the sum of (n-1) terms. This relationship is frequently used in CBSE board problems. |
3.3 Solved Examples on Sum
Example 3.1 — Find the sum of first 20 terms of AP: 5, 8, 11, 14, ... Given: a = 5, d = 3, n = 20 S20 = n/2 × (2a + (n-1)d) S20 = 20/2 × (2×5 + (20-1)×3) S20 = 10 × (10 + 57) S20 = 10 × 67 = 670
Answer: Sum of first 20 terms = 670 |
Example 3.2 — Find the sum: 1 + 3 + 5 + 7 + ... + 99 AP: 1, 3, 5, ..., 99 | a = 1, d = 2, an = 99 Find n: 99 = 1 + (n-1) × 2 => 98 = 2(n-1) => n = 50 Sn = n/2 × (a + l) = 50/2 × (1 + 99) S50 = 25 × 100 = 2500
Answer: Sum of first 50 odd numbers = 2500 |
4. Applications and Special Cases
4.1 Practical Applications of AP
• Finance: Simple interest over multiple years forms an AP.
• Physics: Distance traveled by a uniformly accelerating body forms an AP.
• Architecture: Stairs with equal rise, seats in an auditorium.
• Daily Life: Salary increments, taxi fare meters, installment plans.
4.2 Three Terms in AP
⚑ Choosing 3 Terms in AP When three numbers are in AP, take them as (a - d), a, (a + d). This makes the middle term the average, simplifying equations considerably. |
Example 4.1 — Three numbers in AP have sum 24 and product 480. Find them. Let the three terms be (a - d), a, (a + d) Sum: (a - d) + a + (a + d) = 3a = 24 => a = 8 Product: (8 - d) × 8 × (8 + d) = 480 8(64 - d^2) = 480 => 64 - d^2 = 60 => d^2 = 4 => d = ±2
If d = 2: terms are 6, 8, 10 If d = -2: terms are 10, 8, 6 (same set)
Answer: The three numbers are 6, 8, and 10. |
4.3 Four Terms in AP
When four numbers are in AP, take them as (a - 3d), (a - d), (a + d), (a + 3d). Note that d here represents half the actual common difference, so the actual common difference is 2d.
4.4 Finding Sum When Last Term is Given
Example 4.2 — An AP has first term 5, last term 45, and sum 400. Find n and d. Using Sn = n/2 × (a + l): 400 = n/2 × (5 + 45) = n/2 × 50 = 25n n = 16
Using an = a + (n-1)d: 45 = 5 + (16-1)d => 40 = 15d => d = 8/3
Answer: n = 16, d = 8/3 |
5. Formula Summary Table
All key formulas for Arithmetic Progressions at a glance:
Formula Name | Formula | When to Use |
nth Term (General Term) | an = a + (n-1)d | Find any specific term |
Last Term | l = a + (n-1)d | Find last term of finite AP |
Number of Terms | n = (l - a)/d + 1 | Count terms in AP |
Sum (using d) | Sn = n/2 × [2a + (n-1)d] | Sum when a, d, n known |
Sum (using l) | Sn = n/2 × (a + l) | Sum when first & last term known |
nth Term via Sum | an = Sn - S(n-1) | Find term from sum formula |
Sum of n natural numbers | Sn = n(n+1)/2 | Special case: a=1, d=1 |
Sum of n odd numbers | Sn = n^2 | Special case: a=1, d=2 |
6. Key Theorems and Properties
⚑ Theorem 1 — Arithmetic Mean If three numbers a, b, c are in AP, then b - a = c - b, which gives 2b = a + c. Therefore b = (a + c)/2. The middle term 'b' is called the Arithmetic Mean (AM) of 'a' and 'c'. |
⚑ Theorem 2 — Sum Property In any AP, the sum of terms equidistant from both ends is always equal. That is: a1 + an = a2 + a(n-1) = a3 + a(n-2) = ... = a + l. This property is very useful in quick calculations. |
⚑ Theorem 3 — Constant Difference If each term of an AP is increased, decreased, multiplied, or divided by a constant, the resulting sequence is also an AP. The common difference changes accordingly (addition/subtraction: d unchanged; multiplication: d multiplied by same constant). |
6.1 Properties Quick Reference
Property | Original AP | Modified AP | New Common Difference |
Add constant k | a, a+d, a+2d,... | (a+k), (a+k+d), ... | d (unchanged) |
Subtract constant k | a, a+d, a+2d,... | (a-k), (a-k+d), ... | d (unchanged) |
Multiply by k | a, a+d, a+2d,... | ka, ka+kd, ... | kd |
Divide by k (k≠0) | a, a+d, a+2d,... | a/k, a/k+d/k, ... | d/k |
7. Common Mistakes and Exam Tips
7.1 Frequent Errors to Avoid
1. Forgetting that n must be a positive integer when finding the term number.
2. Using n/2 × (2a + nd) instead of n/2 × (2a + (n-1)d) — off-by-one error.
3. Assuming d = a2 - a1 but then not checking with a3 - a2 to confirm AP.
4. Confusing 'nth term' with 'sum of n terms'.
5. Not setting d correctly when AP is decreasing (d is negative, not the absolute difference).
7.2 Exam Strategy Tips
• Always write down a, d, and n clearly before applying any formula.
• For finding 'which term is x', use an = a + (n-1)d and solve for n.
• If the question asks for 3 terms, assume (a-d), a, (a+d) to save time.
• Verify your answer by substituting back into the original conditions.
• In 5-mark questions, show each step clearly as marks are awarded for working.
8. Practice Questions
Practice the following questions to consolidate your understanding. Questions are organised by marks allocated in CBSE examinations.
8.1 — 1 Mark Questions (MCQ / Very Short Answer)
6. Find the common difference of the AP: 1/3, 5/3, 9/3, 13/3, ...
7. Write the 10th term of the AP: 2, 7, 12, 17, ...
8. Find the number of terms in the AP: 7, 13, 19, ..., 205.
9. Which term of the AP: 21, 18, 15, ... is zero?
10. If the first term of an AP is 5 and the common difference is -3, write the first four terms.
11. Is 184 a term of the AP: 3, 7, 11, 15, ...?
8.2 — 3 Mark Questions (Short Answer)
12. Find the 31st term of an AP whose 11th term is 38 and 16th term is 73.
13. An AP consists of 50 terms of which the 3rd term is 12 and the last term is 106. Find the 29th term.
14. Find the sum of first 15 multiples of 8.
15. How many terms of the AP: 9, 17, 25, ... must be taken to give a sum of 636?
16. The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.
8.3 — 5 Mark Questions (Long Answer)
17. A sum of Rs. 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is Rs. 20 less than its preceding prize, find the value of each of the prizes.
18. In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of Class I will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?
19. Find the sum of all natural numbers between 100 and 200 that are multiples of 3.
20. If the sum of the first n terms of an AP is 4n - n^2, find the first term, the common difference, and the nth term. Also find the 10th term.
21. The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.
CBSE Class 10 Syllabus |
CBSE Class 10 Notes |