Arithmetic Progression
Arithmetic Progression (AP) is one of the most important and widely tested topics in Class 10 mathematics. An arithmetic progression is a sequence of numbers in which each successive term is obtained by adding a fixed number, called the common difference, to the preceding term. This simple yet powerful concept forms the basis for understanding patterns in numbers, sequences, and series. Arithmetic progressions appear naturally in many real-life situations — from the pages of a book and the rungs of a ladder to salary increments, fixed monthly savings, and seating arrangements in auditoriums. In the CBSE Class 10 syllabus, students learn to identify APs, find the common difference, determine the nth term, and calculate the sum of the first n terms. Mastering this topic is essential not only for board examinations but also for competitive exams and higher mathematics, where sequences and series play a foundational role. This comprehensive guide covers definitions, formulas, derivations, types, solved examples, applications, and frequently asked questions about arithmetic progressions.
What is Arithmetic Progression?
An Arithmetic Progression (AP) is a sequence of numbers in which the difference between any two consecutive terms is always the same. This constant difference is called the common difference, denoted by d. The first term of the AP is usually denoted by a (or a1).
Formally, a sequence a1, a2, a3, a4, ... is an arithmetic progression if a2 - a1 = a3 - a2 = a4 - a3 = ... = d for some constant d.
The general form of an AP is: a, a + d, a + 2d, a + 3d, a + 4d, ..., where a is the first term and d is the common difference. Each term is obtained by adding d to the previous term.
The common difference d can be positive (the AP is increasing, e.g., 3, 7, 11, 15, ...), negative (the AP is decreasing, e.g., 20, 17, 14, 11, ...), or zero (all terms are equal, e.g., 5, 5, 5, 5, ...). The common difference is calculated as d = an+1 - an for any two consecutive terms.
An AP can be finite (having a definite number of terms, like the sequence 2, 5, 8, 11, 14) or infinite (continuing indefinitely, like 1, 4, 7, 10, 13, ...). A finite AP has a last term, often denoted by l.
It is important to distinguish between a sequence and a series. A sequence is an ordered list of numbers following a pattern, while a series is the sum of the terms of a sequence. When we add up the terms of an AP, we get an arithmetic series.
Some examples of APs from daily life: the sequence of even numbers (2, 4, 6, 8, ...) with d = 2; the sequence of multiples of 5 (5, 10, 15, 20, ...) with d = 5; the temperature decreasing by 2 degrees each hour (30, 28, 26, 24, ...) with d = -2.
Arithmetic Progression Formula
nth Term (General Term) of an AP:
an = a + (n - 1)d
where a is the first term, d is the common difference, and n is the position of the term.
Sum of First n Terms of an AP:
Sn = (n/2)[2a + (n - 1)d]
Sum when the last term (l) is known:
Sn = (n/2)(a + l)
where l = an = a + (n - 1)d is the last term.
Common Difference:
d = an - an-1 (for any consecutive terms)
nth Term from the End:
nth term from the end = l - (n - 1)d
Arithmetic Mean of Two Numbers:
If a, b, c are in AP, then b = (a + c)/2. Here, b is the arithmetic mean of a and c.
Derivation and Proof
The formulas for the nth term and the sum of n terms of an AP can be derived as follows:
Derivation of the nth Term Formula: an = a + (n - 1)d
Step 1: Consider an AP with first term a and common difference d. The sequence is: a, a + d, a + 2d, a + 3d, ...
Step 2: Observe the pattern:
1st term: a1 = a = a + (1 - 1)d = a + 0d
2nd term: a2 = a + d = a + (2 - 1)d = a + 1d
3rd term: a3 = a + 2d = a + (3 - 1)d = a + 2d
4th term: a4 = a + 3d = a + (4 - 1)d = a + 3d
Step 3: The pattern is clear: for the nth term, the common difference d is added (n - 1) times to the first term a.
Step 4: Therefore, the nth term: an = a + (n - 1)d. This formula allows us to find any term of the AP directly without listing all previous terms.
Derivation of the Sum Formula: Sn = (n/2)[2a + (n - 1)d]
Step 1: Let the AP have n terms: a, a + d, a + 2d, ..., a + (n - 1)d. Let the last term l = a + (n - 1)d.
Step 2: Write the sum forwards: Sn = a + (a + d) + (a + 2d) + ... + (l - d) + l
Step 3: Write the sum backwards: Sn = l + (l - d) + (l - 2d) + ... + (a + d) + a
Step 4: Add these two equations term by term. Each pair of corresponding terms gives (a + l):
2Sn = (a + l) + (a + l) + (a + l) + ... + (a + l) [n times]
Step 5: Therefore: 2Sn = n(a + l), which gives Sn = (n/2)(a + l).
Step 6: Substitute l = a + (n - 1)d: Sn = (n/2)(a + a + (n - 1)d) = (n/2)(2a + (n - 1)d).
This elegant derivation, attributed to the great mathematician Carl Friedrich Gauss (who reportedly discovered it as a schoolboy), shows the power of pairing terms from opposite ends of an AP. The method works because the sum of each pair is constant, making the total sum easy to calculate.
Step 7: Special case: When d = 0 (all terms equal to a), Sn = (n/2)(2a + 0) = na, which simply means n copies of the same number.
Types and Properties
Arithmetic progressions can be classified in several ways based on their properties:
| Type | Common Difference | Behaviour | Example |
|---|---|---|---|
| Increasing AP | d > 0 (positive) | Each term is greater than the previous | 3, 7, 11, 15, 19, ... |
| Decreasing AP | d < 0 (negative) | Each term is less than the previous | 50, 45, 40, 35, 30, ... |
| Constant AP | d = 0 | All terms are equal | 8, 8, 8, 8, 8, ... |
| Finite AP | Any value of d | Has a definite number of terms | 2, 5, 8, 11, 14 (5 terms) |
| Infinite AP | Any value of d | Continues indefinitely | 1, 3, 5, 7, 9, ... |
Important Properties of an AP:
- Property 1: If a constant is added to (or subtracted from) each term of an AP, the resulting sequence is also an AP with the same common difference.
- Property 2: If each term of an AP is multiplied (or divided) by a non-zero constant k, the resulting sequence is also an AP with common difference kd (or d/k).
- Property 3: In a finite AP, the sum of terms equidistant from the beginning and end is constant and equals the sum of the first and last terms: a1 + an = a2 + an-1 = a3 + an-2 = ...
- Property 4: Three numbers a, b, c are in AP if and only if 2b = a + c.
- Property 5: The nth term can also be expressed as an = Sn - Sn-1 for n ≥ 2.
Selecting Terms in AP (Useful for Problem-Solving):
| Number of Terms | Choose Terms As | Why This Works |
|---|---|---|
| 3 terms | a - d, a, a + d | Sum = 3a (d cancels out) |
| 4 terms | a - 3d, a - d, a + d, a + 3d | Sum = 4a (common difference is 2d) |
| 5 terms | a - 2d, a - d, a, a + d, a + 2d | Sum = 5a (d cancels out) |
Solved Examples
Example 1: Checking if a Sequence is an AP
Problem: Check whether the sequence 7, 13, 19, 25, 31, ... is an AP. If yes, find the common difference.
Solution:
Step 1: Find the differences between consecutive terms:
a2 - a1 = 13 - 7 = 6
a3 - a2 = 19 - 13 = 6
a4 - a3 = 25 - 19 = 6
a5 - a4 = 31 - 25 = 6
Step 2: Since the difference is constant (6) for all consecutive pairs, the sequence is an AP.
Answer: Yes, it is an AP with common difference d = 6.
Example 2: Finding the nth Term of an AP
Problem: Find the 20th term of the AP: 5, 8, 11, 14, ...
Solution:
Step 1: Identify a = 5, d = 8 - 5 = 3, n = 20
Step 2: Use the formula an = a + (n - 1)d
Step 3: a20 = 5 + (20 - 1) × 3 = 5 + 19 × 3 = 5 + 57 = 62
Answer: The 20th term is 62.
Example 3: Finding the Number of Terms in an AP
Problem: How many terms are there in the AP: 7, 11, 15, 19, ..., 131?
Solution:
Step 1: Here a = 7, d = 11 - 7 = 4, last term l = 131
Step 2: Using an = a + (n - 1)d: 131 = 7 + (n - 1) × 4
Step 3: 131 - 7 = (n - 1) × 4 → 124 = (n - 1) × 4
Step 4: n - 1 = 124/4 = 31, so n = 32
Answer: There are 32 terms in the AP.
Example 4: Finding the Sum of First n Terms
Problem: Find the sum of the first 15 terms of the AP: 3, 8, 13, 18, ...
Solution:
Step 1: Here a = 3, d = 8 - 3 = 5, n = 15
Step 2: Use Sn = (n/2)[2a + (n - 1)d]
Step 3: S15 = (15/2)[2 × 3 + (15 - 1) × 5] = (15/2)[6 + 70] = (15/2) × 76 = 15 × 38 = 570
Answer: The sum of the first 15 terms is 570.
Example 5: Finding the Sum When First and Last Terms Are Known
Problem: Find the sum of the AP: 2, 7, 12, 17, ..., 152.
Solution:
Step 1: Here a = 2, d = 5, l = 152
Step 2: First find n: 152 = 2 + (n - 1) × 5 → 150 = 5(n - 1) → n - 1 = 30 → n = 31
Step 3: Sn = (n/2)(a + l) = (31/2)(2 + 152) = (31/2) × 154 = 31 × 77 = 2387
Answer: The sum of the AP is 2387.
Example 6: Finding Three Numbers in AP Given Their Sum and Product
Problem: The sum of three numbers in AP is 27 and their product is 648. Find the numbers.
Solution:
Step 1: Let the three numbers be (a - d), a, (a + d).
Step 2: Sum = (a - d) + a + (a + d) = 3a = 27, so a = 9.
Step 3: Product = (a - d)(a)(a + d) = a(a2 - d2) = 9(81 - d2) = 648
Step 4: 81 - d2 = 72, so d2 = 9, giving d = 3 or d = -3.
Step 5: If d = 3: numbers are 6, 9, 12. If d = -3: numbers are 12, 9, 6.
Answer: The three numbers are 6, 9, and 12.
Example 7: Finding a Term Given Two Other Terms
Problem: The 7th term of an AP is 32 and the 13th term is 62. Find the AP and its 20th term.
Solution:
Step 1: a7 = a + 6d = 32 ... (i)
Step 2: a13 = a + 12d = 62 ... (ii)
Step 3: Subtract (i) from (ii): 6d = 30, so d = 5
Step 4: Substitute in (i): a + 30 = 32, so a = 2
Step 5: The AP is 2, 7, 12, 17, 22, ...
Step 6: a20 = 2 + 19 × 5 = 2 + 95 = 97
Answer: The AP is 2, 7, 12, 17, ... and the 20th term is 97.
Example 8: Sum of Natural Numbers Using AP
Problem: Find the sum of the first 100 natural numbers.
Solution:
Step 1: The natural numbers 1, 2, 3, ..., 100 form an AP with a = 1, d = 1, n = 100, l = 100.
Step 2: Sn = (n/2)(a + l) = (100/2)(1 + 100) = 50 × 101 = 5050
Answer: The sum of the first 100 natural numbers is 5050.
Example 9: Word Problem: Saving Money in AP
Problem: Priya saves Rs. 100 in the first month and increases her saving by Rs. 50 every month. How much will she save in 2 years? Also find her savings in the 24th month.
Solution:
Step 1: Monthly savings form an AP: 100, 150, 200, 250, ... with a = 100, d = 50.
Step 2: Number of months in 2 years: n = 24
Step 3: Saving in the 24th month: a24 = 100 + (24 - 1) × 50 = 100 + 1150 = 1250
Step 4: Total savings: S24 = (24/2)[2 × 100 + 23 × 50] = 12[200 + 1150] = 12 × 1350 = 16200
Answer: Priya saves Rs. 1,250 in the 24th month and Rs. 16,200 in total over 2 years.
Example 10: Word Problem: Rows of Seats in an Auditorium
Problem: An auditorium has 20 rows of seats. The first row has 18 seats and each subsequent row has 2 more seats than the previous row. How many seats are there in the last row, and how many seats in total?
Solution:
Step 1: The number of seats per row forms an AP: 18, 20, 22, ... with a = 18, d = 2, n = 20.
Step 2: Seats in the last (20th) row: a20 = 18 + (20 - 1) × 2 = 18 + 38 = 56
Step 3: Total seats: S20 = (20/2)(18 + 56) = 10 × 74 = 740
Answer: The last row has 56 seats, and the auditorium has 740 seats in total.
Real-World Applications
Arithmetic progressions have extensive applications in mathematics, science, business, and everyday life:
Finance and Banking: Fixed monthly savings plans, where the same amount is saved each month (or with a fixed increment), follow arithmetic progressions. Simple interest calculations result in amounts that form an AP over equal time periods. Loan repayment schedules with equal principal installments create an AP of decreasing interest payments.
Physics and Motion: An object moving with uniform acceleration covers distances in successive equal time intervals that form an AP. For example, a freely falling body covers distances of g/2, 3g/2, 5g/2, ... metres in the 1st, 2nd, 3rd seconds, forming an AP with d = g. The concept is fundamental to kinematics and uniformly accelerated motion.
Construction and Architecture: In building design, if each floor of a building has a fixed number of additional windows or rooms compared to the floor below, the number of rooms per floor forms an AP. Pyramid-like structures with decreasing layers often follow AP patterns.
Calendar and Time Patterns: Dates that fall on the same day of the week form an AP with d = 7. Clock chimes that increase by one each hour from 1 to 12 form an AP. Many scheduling and timetable problems use AP concepts.
Daily Life: The rungs of a ladder, if equally spaced, represent positions in an AP. The number of logs in a pile arranged in rows with one fewer log per row forms an AP. Tax brackets with equal increments, regular salary hikes, and pricing structures in business often follow arithmetic progression patterns.
Key Points to Remember
- An Arithmetic Progression (AP) is a sequence where each term after the first is obtained by adding a constant (common difference d) to the previous term.
- The general form of an AP is: a, a + d, a + 2d, a + 3d, ...
- The common difference d = an+1 - an can be positive, negative, or zero.
- The nth term formula is an = a + (n - 1)d, where a is the first term and n is the position.
- The sum of the first n terms is Sn = (n/2)[2a + (n - 1)d] or Sn = (n/2)(a + l) when the last term l is known.
- The nth term from the end of a finite AP is l - (n - 1)d, where l is the last term.
- If three numbers a, b, c are in AP, then b = (a + c)/2 (b is the arithmetic mean of a and c).
- The sum of terms equidistant from the beginning and end of a finite AP is constant: ak + an-k+1 = a + l.
- A quadratic polynomial in n for Sn always represents the sum of an AP (the coefficient of n2 equals d/2).
- To solve problems involving three terms in AP, take them as (a - d), a, (a + d) to simplify calculations.
Practice Problems
- Write the first five terms of the AP whose first term is 10 and common difference is -3.
- Find the 30th term of the AP: 4, 9, 14, 19, ...
- Which term of the AP 21, 18, 15, ... is -81?
- Find the sum of the first 25 terms of the AP: 1, 4, 7, 10, ...
- The 4th term of an AP is 11 and the 8th term is 23. Find the AP, its first term, and common difference.
- Find the sum of all two-digit odd numbers.
- The sum of the first n terms of an AP is 3n² + 5n. Find the AP and its 15th term.
- How many terms of the AP 24, 21, 18, ... must be taken so that the sum is 78?
Frequently Asked Questions
Q1. What is an arithmetic progression (AP)?
An arithmetic progression (AP) is a sequence of numbers in which the difference between any two consecutive terms is always the same. This constant difference is called the common difference (d). For example, 2, 5, 8, 11, 14, ... is an AP with common difference 3.
Q2. What is the formula for the nth term of an AP?
The nth term of an AP is given by aₙ = a + (n - 1)d, where a is the first term, d is the common difference, and n is the position of the term. This formula allows you to find any term directly without listing all previous terms.
Q3. How do you find the sum of n terms of an AP?
The sum of the first n terms of an AP is Sₙ = (n/2)[2a + (n - 1)d], where a is the first term and d is the common difference. If the last term l is known, you can use the simpler formula Sₙ = (n/2)(a + l).
Q4. What is the common difference in an AP?
The common difference (d) is the fixed number added to each term to get the next term in an AP. It is calculated as d = aₙ₊₁ - aₙ (any term minus the previous term). It can be positive (increasing AP), negative (decreasing AP), or zero (constant sequence).
Q5. How do you check if a sequence is an AP?
To check if a sequence is an AP, calculate the differences between all pairs of consecutive terms. If all these differences are equal, the sequence is an AP. Alternatively, for three consecutive terms a, b, c: the sequence is in AP if and only if 2b = a + c.
Q6. What is the sum of the first 100 natural numbers?
The first 100 natural numbers (1, 2, 3, ..., 100) form an AP with a = 1, d = 1, n = 100, l = 100. Using Sₙ = (n/2)(a + l) = (100/2)(1 + 100) = 50 × 101 = 5050. This was famously calculated by Gauss as a young student.
Q7. What is the arithmetic mean of two numbers?
The arithmetic mean of two numbers a and c is (a + c)/2. If three numbers a, b, c are in AP, then b is the arithmetic mean of a and c. For example, the arithmetic mean of 10 and 20 is (10 + 20)/2 = 15, and 10, 15, 20 are in AP.
Q8. Can the common difference of an AP be negative?
Yes, the common difference can be negative. When d < 0, each term is smaller than the previous one, and the AP is a decreasing sequence. For example, 30, 25, 20, 15, 10, ... is an AP with d = -5. The terms decrease by 5 each time.
Q9. How do you find the number of terms in a finite AP?
Use the nth term formula: l = a + (n - 1)d, where l is the last term. Solving for n gives n = (l - a)/d + 1. For example, in the AP 5, 9, 13, ..., 101: n = (101 - 5)/4 + 1 = 96/4 + 1 = 24 + 1 = 25 terms.
Q10. What is the difference between AP and GP?
In an Arithmetic Progression (AP), the difference between consecutive terms is constant (e.g., 3, 7, 11, 15 with d = 4). In a Geometric Progression (GP), the ratio between consecutive terms is constant (e.g., 3, 6, 12, 24 with r = 2). AP involves addition/subtraction, while GP involves multiplication/division.
