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Updated on 22 Jul 2026, 13:27 IST
Arithmetic Progressions Class 10 Notes PDF covers the meaning of an AP, the nth-term formula, the sum of the first n terms, solved examples and exam-style questions. Use these notes to understand each formula, choose the correct method, solve word problems and revise NCERT Class 10 Maths Chapter 5.
The current CBSE Class 10 Mathematics curriculum covers the motivation for studying arithmetic progressions, derivation of the nth-term and sum formulas, and their application to daily-life problems.
Also Check: Class 10 Maths revision notes
Arithmetic Progressions introduces number sequences in which the difference between consecutive terms remains constant. This fixed value is called the common difference and may be positive, negative or zero. The chapter develops methods for recognising an arithmetic progression, finding an unknown term and calculating the sum of a given number of terms.
aₙ = a + (n − 1)d
They also learn the two formulas for finding the sum of the first n terms:
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Sₙ = n/2 [2a + (n − 1)d]
Sₙ = n/2 (a + l)
The chapter includes finding missing terms, determining whether a number belongs to an AP, calculating the number of terms and inserting arithmetic means. It also applies arithmetic progressions to situations involving savings, seating arrangements, production patterns and other quantities that increase or decrease by a fixed amount.
The Arithmetic Progressions Class 10 Notes PDF contains definitions, formulas, derivations, worked examples, common mistakes and graded practice questions.
An arithmetic progression is a sequence in which the difference between every term and the term immediately before it remains constant.
Consider the sequence:
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3, 7, 11, 15, 19, …
Its consecutive differences are:
7 − 3 = 4
11 − 7 = 4
15 − 11 = 4
Because the difference is always 4, this sequence is an arithmetic progression with common difference d = 4.
| Symbol | Meaning | Example |
| a | First term | 3 |
| d | Common difference | 4 |
| n | Position of a term or number of terms | 10 |
| aₙ | Term at position n | a₁₀ |
| l | Last term of a finite AP | Depends on the AP |
| Sₙ | Sum of the first n terms | S₁₀ |
Important: aₙ represents one term, while Sₙ represents the total of the first n terms.
The common difference is calculated by subtracting a term from the term immediately after it:
d = a₂ − a₁
The same difference must occur between every pair of consecutive terms.
| Arithmetic progression | Calculation | Common difference |
| 5, 8, 11, 14, … | 8 − 5 | 3 |
| 18, 13, 8, 3, … | 13 − 18 | −5 |
| 7, 7, 7, 7, … | 7 − 7 | 0 |
| ½, 1, 1½, 2, … | 1 − ½ | ½ |
A decreasing AP has a negative common difference. A constant sequence is also an AP because its common difference is zero.
| Sequence | Consecutive differences | Is it an AP? |
| 2, 5, 8, 11 | 3, 3, 3 | Yes |
| 10, 6, 2, −2 | −4, −4, −4 | Yes |
| 1, 4, 9, 16 | 3, 5, 7 | No |
| 6, 6, 6, 6 | 0, 0, 0 | Yes |
Increasing AP: An arithmetic progression with d > 0, such as 2, 5, 8, 11, …
Decreasing AP: An arithmetic progression with d < 0, such as 20, 16, 12, 8, …
Constant AP: An arithmetic progression with d = 0, such as 9, 9, 9, 9, …
Finite AP: An arithmetic progression with a fixed number of terms, such as 4, 7, 10, 13.
Infinite AP: An arithmetic progression that continues without ending, such as 4, 7, 10, 13, … An infinite AP has an nth term but does not have a last term.
| Term | Meaning | Example |
| Sequence | An ordered list of numbers | 2, 4, 6, 8 |
| Arithmetic progression | A sequence with a constant difference | 2, 4, 6, 8 |
| Series | The result of adding sequence terms | 2 + 4 + 6 + 8 |
The main Arithmetic Progressions Class 10 formulas calculate a particular term, the number of terms and the sum of the first n terms.
| What you need to find | Formula |
| Common difference | d = aₙ − aₙ₋₁ |
| nth term | aₙ = a + (n − 1)d |
| Last term | l = a + (n − 1)d |
| Number of terms | n = (l − a)/d + 1 |
| Sum when a, d and n are known | Sₙ = n/2 [2a + (n − 1)d] |
| Sum when a, l and n are known | Sₙ = n/2 (a + l) |
| nth term when sums are known | aₙ = Sₙ − Sₙ₋₁ |
The value of n must be a positive integer when it represents a term position or the number of terms.
| Information given | What must be found | Best method |
| a, d and n | One particular term | aₙ = a + (n − 1)d |
| a, d and n | Total of the terms | Sₙ = n/2 [2a + (n − 1)d] |
| a, l and n | Total of the terms | Sₙ = n/2 (a + l) |
| Two specified terms | a and d | Form two nth-term equations |
| a, d and l | Number of terms | Rearrange the nth-term formula |
| A formula for Sₙ | aₙ | aₙ = Sₙ − Sₙ₋₁ |
Also Check: Class 10 Maths formula sheet
The nth term of an arithmetic progression is calculated using aₙ = a + (n − 1)d, where a is the first term and d is the common difference.
Consider an AP with first term a and common difference d:
a, a + d, a + 2d, a + 3d, …
Its terms can be written as:
a₁ = a
a₂ = a + d
a₃ = a + 2d
a₄ = a + 3d
The fourth term contains three additions of d. In the same way, the nth term contains n − 1 additions of d.
Therefore:
aₙ = a + (n − 1)d
The formula does not use a + nd because there are only n − 1 jumps from the first term to the nth term.
Find the 20th term of:
5, 9, 13, 17, …
Given:
a = 5
d = 9 − 5 = 4
n = 20
Apply the nth-term formula:
a₂₀ = a + (20 − 1)d
a₂₀ = 5 + 19 × 4
a₂₀ = 5 + 76
a₂₀ = 81
Answer: The 20th term is 81.
Which term of 7, 12, 17, 22, … is 157?
Given:
a = 7
d = 5
aₙ = 157
Substitute the values:
157 = 7 + (n − 1)5
150 = 5(n − 1)
30 = n − 1
n = 31
Answer: 157 is the 31st term.
Is 100 a term of 4, 10, 16, 22, …?
Given:
a = 4
d = 6
aₙ = 100
100 = 4 + (n − 1)6
96 = 6(n − 1)
16 = n − 1
n = 17
Answer: Yes. Because n = 17 is a positive integer, 100 is the 17th term.
Is 50 a term of 3, 8, 13, 18, …?
Given:
a = 3
d = 5
aₙ = 50
50 = 3 + (n − 1)5
47 = 5(n − 1)
n − 1 = 9.4
n = 10.4
A term position cannot be 10.4.
Answer: 50 is not a term of this AP.
If the last term l is known:
l = a + (n − 1)d
Rearranging the formula gives:
n = (l − a)/d + 1
Find the number of terms in:
8, 13, 18, …, 148
Given:
a = 8
d = 5
l = 148
n = (148 − 8)/5 + 1
n = 140/5 + 1
n = 28 + 1
n = 29
Answer: The AP contains 29 terms.
The rth term from the end of a finite AP is:
l − (r − 1)d
Find the fifth term from the end of:
3, 7, 11, …, 99
Given:
l = 99
d = 4
r = 5
Fifth term from the end = 99 − (5 − 1)4
= 99 − 16
= 83
Answer: The fifth term from the end is 83.
The sum of the first n terms is calculated using Sₙ = n/2 [2a + (n − 1)d], or Sₙ = n/2 (a + l) when the last term is known.
Write the sum of an AP:
Sₙ = a + (a + d) + (a + 2d) + … + [a + (n − 1)d]
Write the same terms in reverse order:
Sₙ = [a + (n − 1)d] + [a + (n − 2)d] + … + a
Add the two equations. Each pair has the same value, 2a + (n − 1)d, and there are n pairs:
2Sₙ = n[2a + (n − 1)d]
Divide both sides by 2:
Sₙ = n/2 [2a + (n − 1)d]
Because l = a + (n − 1)d, the formula can also be written as:
Sₙ = n/2 (a + l)
Find the sum of the first 25 terms of:
4, 7, 10, 13, …
Given:
a = 4
d = 3
n = 25
S₂₅ = 25/2 [2(4) + (25 − 1)3]
S₂₅ = 25/2 [8 + 72]
S₂₅ = 25/2 × 80
S₂₅ = 1000
Answer: The sum of the first 25 terms is 1000.
Find:
5 + 9 + 13 + … + 101
First find the number of terms:
101 = 5 + (n − 1)4
96 = 4(n − 1)
24 = n − 1
n = 25
Now use the sum formula:
S₂₅ = 25/2 (5 + 101)
S₂₅ = 25/2 × 106
S₂₅ = 1325
Answer: The sum is 1325.
How many terms of 2, 5, 8, … have a sum of 155?
Given:
a = 2
d = 3
Sₙ = 155
155 = n/2 [2(2) + (n − 1)3]
310 = n[4 + 3n − 3]
310 = n(3n + 1)
3n² + n − 310 = 0
(3n + 31)(n − 10) = 0
Therefore:
n = 10 or n = −31/3
The negative value is invalid because the number of terms must be a positive integer.
Answer: The required number of terms is 10.
The nth term equals the difference between the sum of the first n terms and the sum of the first n − 1 terms:
aₙ = Sₙ − Sₙ₋₁
Suppose:
Sₙ = 3n² + 2n
Replace n with n − 1:
Sₙ₋₁ = 3(n − 1)² + 2(n − 1)
Sₙ₋₁ = 3n² − 4n + 1
Now subtract:
aₙ = (3n² + 2n) − (3n² − 4n + 1)
aₙ = 6n − 1
Answer: The nth term is 6n − 1.
Missing values in an AP are found by writing each known term as a + (n − 1)d and solving the resulting equations.
The second term of an AP is 8 and the seventeenth term is 53. Find a and d.
For the second term:
a + d = 8 Equation 1
For the seventeenth term:
a + 16d = 53 Equation 2
Subtract Equation 1 from Equation 2:
15d = 45
d = 3
Substitute d = 3 into Equation 1:
a + 3 = 8
a = 5
Answer: The first term is 5 and the common difference is 3.
Find x if 6, x, 18 are consecutive terms of an AP.
Consecutive differences must be equal:
x − 6 = 18 − x
2x = 24
x = 12
Answer: The missing term is 12.
If k arithmetic means are inserted between two numbers, the completed AP contains k + 2 terms.
Insert four arithmetic means between 3 and 33.
The completed AP contains six terms:
a = 3
l = 33
n = 6
Use the nth-term formula:
33 = 3 + (6 − 1)d
30 = 5d
d = 6
The AP is:
3, 9, 15, 21, 27, 33
Answer: The four arithmetic means are 9, 15, 21 and 27.
| Number of terms | Useful representation |
| Three terms | a − d, a, a + d |
| Four terms | a − 3d, a − d, a + d, a + 3d |
| Five terms | a − 2d, a − d, a, a + d, a + 2d |
These symmetrical forms are useful when the sum of the selected terms is given.
An AP word problem becomes easier when the first value, fixed change and required term or total are identified before selecting a formula.
| Wording in the question | Mathematical meaning |
| The 12th term is 35 | a + 11d = 35 |
| The seventh term exceeds the fifth by 12 | a₇ − a₅ = 12 |
| The amount in the 20th month | Find a₂₀ |
| The total after 20 months | Find S₂₀ |
| The value decreases by 3 each time | d = −3 |
| Insert four terms between two numbers | Form an AP containing six terms |
A hall has 20 seats in its first row. Each following row contains two more seats than the previous row. Find the number of seats in the 15th row.
Given:
a = 20
d = 2
n = 15
The question asks for the seats in one row, so use aₙ:
a₁₅ = 20 + (15 − 1)2
a₁₅ = 20 + 28
a₁₅ = 48
Answer: The 15th row contains 48 seats.
To find the total number of seats in the first 15 rows, use S₁₅:
S₁₅ = 15/2 [2(20) + (15 − 1)2]
S₁₅ = 15/2 (40 + 28)
S₁₅ = 15/2 × 68
S₁₅ = 510
Answer: The first 15 rows contain 510 seats altogether.
“In the 15th row” asks for a₁₅, while “in the first 15 rows” asks for S₁₅.
A student saves ₹100 in the first month and increases the monthly saving by ₹25. Find the amount saved in the 12th month and the total saved during 12 months.
Given:
a = 100
d = 25
n = 12
Amount saved in the 12th month:
a₁₂ = 100 + (12 − 1)25
a₁₂ = 100 + 275
a₁₂ = 375
Total saved during 12 months:
S₁₂ = 12/2 [2(100) + (12 − 1)25]
S₁₂ = 6(200 + 275)
S₁₂ = 6 × 475
S₁₂ = 2850
Answer: The student saves ₹375 in the 12th month and ₹2,850 altogether.
A machine produces 500 units on the first day. Its production decreases by 15 units each day. Find its production on the tenth day.
Given:
a = 500
d = −15
n = 10
a₁₀ = 500 + (10 − 1)(−15)
a₁₀ = 500 − 135
a₁₀ = 365
Answer: The machine produces 365 units on the tenth day.
Important AP Class 10 questions test identification, formula selection, equation formation and real-life applications instead of formula recall alone.
1. What is the common difference of 13, 9, 5, 1, …?
A. 4
B. −4
C. −5
D. 5
Answer: B. −4
2. What is the nth term of 7, 10, 13, …?
A. 3n + 4
B. 3n + 7
C. 7n − 3
D. 3n − 4
Answer: A. 3n + 4
3. Which sequence is not an AP?
A. 2, 4, 6, 8
B. 7, 7, 7, 7
C. 16, 12, 8, 4
D. 1, 4, 9, 16
Answer: D. 1, 4, 9, 16
4. If a = 5, d = 3 and n = 10, what is a₁₀?
A. 32
B. 35
C. 30
D. 38
Answer: A. 32
5. If Sₙ = 5n², what is aₙ?
A. 5n
B. 10n − 5
C. 10n + 5
D. 5n − 5
Answer: B. 10n − 5
Assertion: The sequence 8, 8, 8, 8, … is an AP.
Reason: Its common difference is zero.
Answer: Both statements are true, and the reason correctly explains the assertion.
A staircase display uses 6 lights on the first step, 10 on the second, 14 on the third and so on.
Solution:
d = 10 − 6 = 4
a₁₈ = 6 + 17 × 4
a₁₈ = 74
S₁₈ = 18/2 [2(6) + 17(4)]
S₁₈ = 9(12 + 68)
S₁₈ = 720
Answer: The 18th step contains 74 lights, while all 18 steps contain 720 lights. The first value represents one term; the second represents a cumulative total.
CBSE’s competency-based assessment resources include questions that assess interpretation, method selection and mathematical application.
The most frequent AP errors are using n instead of n − 1, ignoring a negative common difference and confusing aₙ with Sₙ.
| Mistake | Why it is incorrect | Correct method |
| Writing aₙ = a + nd | There are only n − 1 jumps after the first term | Use aₙ = a + (n − 1)d |
| Making a negative d positive | A decreasing AP has d < 0 | Calculate later term minus earlier term |
| Using Sₙ for one term | Sₙ is a cumulative total | Use aₙ |
| Accepting a fractional n | A term position must be a positive integer | The proposed value is not a term |
| Saying an infinite AP has no nth term | It has no last term, but it has an nth term | Use aₙ for any positive integer n |
| Confusing l with the numeral 1 | The symbols have different meanings | Use consistent notation |
Incorrect solution:
a = 4, d = 3 and a₁₀ = 4 + 10 × 3 = 34
Error: The calculation uses n instead of n − 1.
Correction:
a₁₀ = 4 + (10 − 1)3
a₁₀ = 31
Incorrect solution:
For 20, 15, 10, 5, …, a student writes d = 5.
Error: The subtraction was performed in the wrong order.
Correction:
d = 15 − 20
d = −5
Incorrect solution:
A student uses S₁₂ to find the amount deposited only in the 12th month.
Error: S₁₂ gives the total deposited during all 12 months.
Correction: Use a₁₂ to find the amount deposited in the 12th month.
An arithmetic progression has a constant difference, whereas a geometric progression has a constant ratio.
| Feature | Arithmetic progression | Geometric progression |
| Constant relationship | Difference | Ratio |
| Main operation | Addition or subtraction | Multiplication or division |
| Example | 2, 5, 8, 11 | 2, 6, 18, 54 |
| Pattern type | Additive | Multiplicative |
Geometric progression formulas are outside the scope of these Class 10 Arithmetic Progressions notes.
Arithmetic Progressions can be revised efficiently by learning the notation, understanding formula selection and completing mixed questions without using notes.
Memorise:
Understand:
NCERT exercises should form the foundation of AP Class 10 preparation because they follow the prescribed chapter concepts and terminology. Students should also practise current official sample questions and competency-focused problems to improve application and interpretation skills.
Also Check: NCERT Class 10 Maths solutions
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An arithmetic progression is a sequence in which the difference between consecutive terms remains constant. For example, 5, 8, 11, 14, … is an AP with d = 3.
The main formulas are aₙ = a + (n − 1)d, Sₙ = n/2 [2a + (n − 1)d] and Sₙ = n/2 (a + l). The correct formula depends on whether the question asks for one term or the sum of several terms.
Subtract a term from the term immediately after it: d = a₂ − a₁. Check at least one more pair to confirm that the difference is constant.
Use aₙ = a + (n − 1)d. Substitute the first term, common difference and required position into the formula.
Use Sₙ = n/2 [2a + (n − 1)d] when the first term and common difference are known. Use Sₙ = n/2 (a + l) when the first term, last term and number of terms are known.
Set the number equal to a + (n − 1)d and solve for n. The number belongs to the AP only if the result is a positive integer.
Write an nth-term equation for each known term and subtract the two equations. The first term is eliminated, allowing the common difference to be calculated.
aₙ is the value of the single term at position n. Sₙ is the sum of every term from the first term through the nth term.
Yes. A decreasing AP has a negative common difference, while a constant AP has a common difference of zero.
Yes. An infinite AP has a term for every positive integer position, but it does not have a last term.
Yes. The CBSE Class 10 Mathematics curriculum includes the nth term, sum of the first n terms and applications of arithmetic progressions.
Yes. Both courses require the central definitions and formulas. Standard Mathematics preparation should include additional reverse, multi-step and equation-based problems.
Download Arithmetic Progressions Class 10 Notes from Infinity Learn websites. A separate one-page formula sheet and practice worksheet should also be provided.