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Arithmetic Progressions

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Chapter Notes

Overview
This chapter introduces Arithmetic Progressions (AP), a sequence of numbers where each term after the first is obtained by adding a fixed number (common difference) to the preceding term. It exists in the syllabus to develop pattern recognition and algebraic reasoning, enabling students to solve real-life problems involving regular increments or decrements. The one big idea a student must walk away with is that any AP can be fully described by its first term (a) and common difference (d), and that the nth term and sum of n terms can be found using simple formulas.
Key Concepts (10)
Arithmetic Progression (AP)
A list of numbers in which each term is obtained by adding a fixed number to the preceding term except the first term.
e.g. The sequence 8000, 8500, 9000, 9500, ... is an AP because each term is obtained by adding 500 to the previous term.
Common difference (d)
The fixed number that is added to each term to get the next term in an AP. It can be positive, negative, or zero.
e.g. In the AP 21, 18, 15, 12, ..., the common difference is 18 - 21 = -3.
First term (a)
The first number in an arithmetic progression, denoted by a or a₁.
e.g. In the AP 2, 7, 12, 17, ..., the first term a = 2.
nth term (aₙ)
The term at the nth position in an AP, given by the formula aₙ = a + (n - 1)d.
e.g. For the AP 3, 5, 7, 9, ..., the 10th term is 3 + (10 - 1)×2 = 21.
General form of an AP
The representation of an AP as a, a + d, a + 2d, a + 3d, ...
e.g. If a = 6 and d = 3, the general form is 6, 9, 12, 15, ...
Finite AP
An AP that has a finite number of terms and a last term.
e.g. The heights of students 147, 148, 149, ..., 157 is a finite AP.
Infinite AP
An AP that does not have a last term and continues indefinitely.
e.g. The sequence 1, 2, 3, 4, ... is an infinite AP.
Last term (l)
The final term of a finite AP, often denoted by l or aₘ.
e.g. In the AP 10, 7, 4, ..., -62, the last term l = -62.
Sum of first n terms (Sₙ)
The total of the first n terms of an AP, given by Sₙ = n/2 [2a + (n - 1)d] or Sₙ = n/2 (a + l).
e.g. For the AP 100, 150, 200, ..., the sum of first 21 terms is ₹12600.
Simple Interest (SI)
Interest calculated using the formula SI = (P × R × T)/100, where P is principal, R is rate, and T is time.
e.g. For ₹1000 at 8% per annum, SI after 1 year is ₹80, after 2 years ₹160, forming an AP.
Formulas & Equations (6)
aₙ = a + (n - 1)d
Reads as: a sub n equals a plus (n minus 1) times d
Variables: aₙ = nth term, a = first term, n = term number (positive integer), d = common difference
Use when: To find any specific term in an AP when first term and common difference are known.
⚠ Students often write aₙ = a + (n + 1)d or forget to subtract 1 from n, leading to incorrect term values.
d = aₖ₊₁ - aₖ
Reads as: d equals a sub k plus 1 minus a sub k
Variables: d = common difference, aₖ₊₁ = (k+1)th term, aₖ = kth term
Use when: To find the common difference from any two consecutive terms of an AP.
⚠ Students subtract the larger term from the smaller term incorrectly; always subtract the kth term from the (k+1)th term, even if the (k+1)th term is smaller.
Sₙ = n/2 [2a + (n - 1)d]
Reads as: S sub n equals n over 2 times [2a plus (n minus 1) times d]
Variables: Sₙ = sum of first n terms, a = first term, n = number of terms, d = common difference
Use when: To find the sum of the first n terms when first term and common difference are given.
⚠ Students forget to multiply d by (n-1) correctly or misplace brackets, e.g., writing n/2 × 2a + (n-1)d instead of n/2 [2a + (n-1)d].
Sₙ = n/2 (a + l)
Reads as: S sub n equals n over 2 times (a plus l)
Variables: Sₙ = sum of first n terms, a = first term, l = last term, n = number of terms
Use when: To find the sum when the first and last terms are known and common difference is not needed.
⚠ Students use this formula even when the last term is not given, or they forget that l = aₙ.
aₙ = Sₙ - Sₙ₋₁
Reads as: a sub n equals S sub n minus S sub n minus 1
Variables: aₙ = nth term, Sₙ = sum of first n terms, Sₙ₋₁ = sum of first (n-1) terms
Use when: To find the nth term when the sum of n terms and sum of (n-1) terms are known.
⚠ Students confuse Sₙ with aₙ or forget that Sₙ₋₁ is the sum of (n-1) terms, not n-1.
Simple Interest = (P × R × T)/100
Reads as: Simple Interest equals P times R times T divided by 100
Variables: P = principal amount (₹), R = rate of interest per annum (%), T = time (years)
Use when: To calculate simple interest for each year, which forms an AP when T increases by 1 each year.
⚠ Students forget to divide by 100 or use the formula for compound interest instead.
How It Works
Identifying an Arithmetic Progression
  1. Step 1: List the given numbers in order.
  2. Step 2: Calculate the difference between each consecutive pair: a₂ - a₁, a₃ - a₂, a₄ - a₃, etc.
  3. Step 3: Check if all these differences are equal.
  4. Step 4: If they are equal, the list forms an AP with common difference d equal to that constant difference.
💡 You only need to check one pair of consecutive terms to find d, but you must verify at least two differences to confirm it is an AP.
Finding the nth term of an AP
  1. Step 1: Identify the first term (a) and common difference (d) from the given AP.
  2. Step 2: Write the formula aₙ = a + (n - 1)d.
  3. Step 3: Substitute the values of a, d, and the required term number n.
  4. Step 4: Simplify to find aₙ.
💡 Always subtract 1 from n before multiplying by d; a common error is using n instead of (n-1).
Finding the sum of first n terms of an AP
  1. Step 1: Identify a, d, and n from the problem.
  2. Step 2: Use the formula Sₙ = n/2 [2a + (n - 1)d].
  3. Step 3: Substitute the values and simplify step-by-step.
  4. Step 4: Alternatively, if the last term l is known, use Sₙ = n/2 (a + l).
💡 The sum formula gives the total of the first n terms, not the nth term. Double-check whether you need Sₙ or aₙ.
Finding the number of terms in an AP given the sum
  1. Step 1: Write the sum formula Sₙ = n/2 [2a + (n - 1)d].
  2. Step 2: Substitute the known values of Sₙ, a, and d.
  3. Step 3: Simplify to form a quadratic equation in n.
  4. Step 4: Solve the quadratic equation (e.g., by factoring) to find n.
  5. Step 5: Reject any non-positive or non-integer values of n.
💡 Sometimes two values of n are possible (e.g., n = 4 and n = 13) because some terms cancel out. Both are valid unless the context says otherwise.
Finding the 11th term from the last term of an AP
  1. Step 1: Find the total number of terms (n) in the AP using aₙ = a + (n - 1)d.
  2. Step 2: The kth term from the last is the (n - k + 1)th term from the beginning.
  3. Step 3: Use aₙ formula with the new term number to find the required term.
  4. Step 4: Alternatively, reverse the AP (make last term first, reverse sign of d) and find the kth term directly.
💡 The 11th term from the last is the 15th term if there are 25 terms, not the 14th. Always calculate n first.
Exam Traps — Don't Make These Mistakes
⚠ Confusing nth term with sum of n terms
✗ They write Sₙ = a + (n - 1)d instead of aₙ = a + (n - 1)d.
→ Sₙ is the sum of the first n terms, not the nth term. Using the wrong formula gives a completely different value.
✓ Write aₙ = a + (n - 1)d for the nth term, and Sₙ = n/2 [2a + (n - 1)d] for the sum.
⚠ Incorrect sign when finding common difference
✗ They write d = a₁ - a₂ instead of d = a₂ - a₁.
→ The common difference is defined as the term minus the preceding term, so d = a₂ - a₁. Reversing the order gives the wrong sign.
✓ Always subtract the earlier term from the later term: d = aₖ₊₁ - aₖ.
⚠ Forgetting to subtract 1 in nth term formula
✗ They write aₙ = a + n d instead of aₙ = a + (n - 1)d.
→ The first term is a + 0d, so the nth term should have (n-1) times d. Using n d gives the (n+1)th term.
✓ Write aₙ = a + (n - 1)d and always check for n=1: a₁ should equal a.
⚠ Using sum formula with last term when not given
✗ They use Sₙ = n/2 (a + l) without knowing l, or they assume l = aₙ incorrectly.
→ The formula Sₙ = n/2 (a + l) requires the actual last term l. If l is not given, use Sₙ = n/2 [2a + (n - 1)d].
✓ First check if l is provided. If yes, use Sₙ = n/2 (a + l). If not, use the formula with d.
⚠ Misinterpreting 'term from the last'
✗ They think the 11th term from the last is the 11th term of the original AP.
→ The 11th term from the last is the (n - 10)th term from the beginning, not the 11th term. For example, in a 25-term AP, the 11th from last is the 15th term.
✓ Find total n, then the required term number = n - k + 1, where k is the position from the last.
⚠ Not checking if a number is a term of an AP
✗ They solve a + (n - 1)d = given number and accept any n, even if n is not a positive integer.
→ n must be a positive integer (1, 2, 3, ...). If solving gives a fraction or negative number, the given number is not a term of the AP.
✓ After solving, check if n is a positive integer. If not, state that the number is not a term of the AP.
Compare & Contrast
Pattern of progression
Arithmetic Progression (AP)
Geometric Progression (GP)
In an AP, each term is obtained by adding a fixed number (common difference) to the previous term. In a GP, each term is obtained by multiplying the previous term by a fixed number (common ratio).
Number of terms
Finite AP
Infinite AP
A finite AP has a limited number of terms and a last term, e.g., 147, 148, ..., 157. An infinite AP continues indefinitely without a last term, e.g., 1, 2, 3, 4, ...
Exam Focus
Marks pattern: Typically 1 MCQ (1 mark), 1 SA-I (2 marks), 1 SA-II (3 marks), and 1 LA (5 marks) — total around 11 marks.
Most asked: Finding the nth term of an AP · Finding the sum of first n terms of an AP · Word problems involving AP (e.g., salary, savings, production) · Determining if a given list forms an AP
Don't skip: Word problems on sum of n terms (e.g., total production, penalty, potato race); Finding number of terms given sum (quadratic equation in n); Finding term from the last
One-Line Revision (read aloud before the exam)
  1. An arithmetic progression is a sequence where each term after the first is obtained by adding a fixed number called the common difference.
  2. The common difference d can be positive, negative, or zero.
  3. The general form of an AP is a, a + d, a + 2d, a + 3d, ...
  4. The nth term of an AP is given by aₙ = a + (n - 1)d.
  5. The sum of the first n terms of an AP is Sₙ = n/2 [2a + (n - 1)d].
  6. If the first and last terms are known, use Sₙ = n/2 (a + l).
  7. The common difference d is found by subtracting any term from its succeeding term: d = aₖ₊₁ - aₖ.
  8. To check if a number is a term of an AP, solve a + (n - 1)d = number and see if n is a positive integer.
  9. The kth term from the last of an AP with n terms is the (n - k + 1)th term from the beginning.
  10. The nth term can also be found as aₙ = Sₙ - Sₙ₋₁.
  11. Simple interest for successive years forms an AP with common difference = (P × R)/100.
  12. In word problems, identify the first term and common difference from the context before applying formulas.

MCQs — 10 Questions with Answers

Q1. A student observes that in the sequence 8000, 8500, 9000, 9500, ..., each term is obtained by adding 500 to the previous term. Which property of Arithmetic Progressions does this directly illustrate?

  • A. The common difference is constant between consecutive terms ✓
  • B. The first term determines whether the AP is finite or infinite
  • C. All APs must have a positive common difference
  • D. The AP can be used to model only salary increments

Answer: A — The definition of an AP requires that the difference between consecutive terms (common difference d) is constant; in this case d = 500, which is exactly what distinguishes an AP from other sequences.

Q2. Consider the ladder rung lengths: 45, 43, 41, 39, 37, 35, 33, 31. A student claims this is not an AP because the lengths are decreasing. Using the definition of an AP, is this claim correct?

  • A. Yes, the claim is correct because decreasing sequences cannot be APs
  • B. No, the claim is incorrect; this is an AP with a negative common difference of –2 ✓
  • C. No, the claim is incorrect; the sequence is AP only if all terms are positive
  • D. Yes, the claim is correct because APs must have a positive common difference

Answer: B — An AP is defined by a constant common difference (here d = –2), not by whether the sequence increases or decreases; negative common differences are explicitly allowed by the AP definition.

Q3. Assertion (A): The sequence 3, 3, 3, 3, ... is an Arithmetic Progression. Reason (R): In this sequence, the common difference d = 0, and the definition of AP permits d to be zero. Choose the correct option:

  • A. Both A and R are true and R is the correct explanation of A ✓
  • B. Both A and R are true but R is not the correct explanation of A
  • C. A is true but R is false
  • D. A is false but R is true

Answer: A — Both statements are true; R directly explains why A is true because the definition explicitly states d can be any fixed number including zero.

Q4. A student is given the sequence –1.0, –1.5, –2.0, –2.5, ... and claims it cannot be an AP because it contains decimals and negative numbers. Based on the AP definition, evaluate this reasoning.

  • A. The student is correct; APs are defined only for positive integers
  • B. The student is incorrect; APs can have any real numbers as terms and the common difference is d = –0.5 ✓
  • C. The student is correct; the presence of decimals makes the sequence irregular
  • D. The student is incorrect, but only because the common difference is positive

Answer: B — The AP definition makes no restriction on the type of numbers (integers, decimals, negative numbers are all allowed); here d = –0.5 is constant, confirming it is an AP.

Q5. Assertion (A): To completely determine an AP, you must know both the first term a and the common difference d. Reason (R): Knowing only a without d, or only d without a, does not allow you to write down the sequence uniquely. Choose the correct option:

  • A. Both A and R are true and R is the correct explanation of A ✓
  • B. Both A and R are true but R is not the correct explanation of A
  • C. A is true but R is false
  • D. A is false but R is true

Answer: A — R directly justifies A: without both parameters, different APs are possible (e.g., a = 6, d = 3 gives 6, 9, 12, ... while a = 6, d = –3 gives 6, 3, 0, ..., both different despite sharing one parameter).

Q6. A teacher writes two sequences on the board: Sequence P: 100, 150, 200, 250, ... and Sequence Q: 12, 22, 32, .... The teacher claims both are APs. Based on the AP definition, which claim is correct?

  • A. Both sequences are APs because both have a starting number
  • B. Only Sequence P is an AP; Sequence Q is not because it contains squares ✓
  • C. Both are APs; the form of the terms does not matter, only the constant difference
  • D. Neither is an AP because APs must decrease

Answer: B — Sequence P has a constant common difference d = 50, making it an AP; Sequence Q (1, 4, 9, 16, ...) has differences 3, 5, 7, ... which are not constant, so it is not an AP.

Q7. For the general form of an AP: a, a + d, a + 2d, a + 3d, ..., a student writes the nth term as a + (n – 1)d instead of a + nd. Is this correct or incorrect?

  • A. Incorrect; the formula should always be a + nd
  • B. Correct; a + (n – 1)d gives the nth term because the first term is a (which is a + 0·d) ✓
  • C. Incorrect; the formula depends on whether d is positive or negative
  • D. Correct only for finite APs with a fixed number of terms

Answer: B — The nth term formula a + (n – 1)d is correct because the first term (n = 1) is a + (1 – 1)d = a, the second term (n = 2) is a + (2 – 1)d = a + d, confirming the pattern.

Q8. Assertion (A): The list of heights 147, 148, 149, ..., 157 is a finite Arithmetic Progression. Reason (R): A finite AP has a last term, whereas infinite APs do not have a last term. Choose the correct option:

  • A. Both A and R are true and R is the correct explanation of A ✓
  • B. Both A and R are true but R is not the correct explanation of A
  • C. A is true but R is false
  • D. A is false but R is true

Answer: A — R directly explains A: the sequence has a clear last term (157) and constant difference (1), making it a finite AP by definition.

Q9. A student claims that the sequence 6, 9, 12, 15, ... proves that knowing only the common difference d = 3 is enough to determine an AP uniquely. Why is this claim incomplete?

  • A. Because the common difference must be checked at every term, not just once
  • B. Because d alone cannot determine the AP; the first term a is also essential (here a = 6), and different values of a give different APs ✓
  • C. Because this AP is infinite, and infinite APs require additional information
  • D. Because d = 3 is positive, and we would need to test negative differences too

Answer: B — Knowing d = 3 alone leaves infinitely many possible APs (3, 6, 9, ..., or 10, 13, 16, ..., etc.); both a and d are needed to uniquely determine the AP.

Q10. Assertion (A): The balance money sequence 950, 900, 850, 800, ..., 50 is an Arithmetic Progression. Reason (R): The common difference is –50, and each term is obtained by subtracting 50 from the previous term. Choose the correct option:

  • A. Both A and R are true and R is the correct explanation of A ✓
  • B. Both A and R are true but R is not the correct explanation of A
  • C. A is true but R is false
  • D. A is false but R is true

Answer: A — R directly justifies A: the constant difference d = –50 between consecutive terms is the defining property of an AP.

Flashcards

What is an Arithmetic Progression (AP)?

A sequence where each term is obtained by adding a fixed number (common difference d) to the preceding term.

What is the common difference d in an AP?

The fixed number added to each term to get the next term, found by subtracting any term from the term immediately following it.

Write the general form of an AP.

a, a+d, a+2d, a+3d, ... where a is the first term and d is the common difference.

Formula for the nth term of an AP

an = a + (n-1)d, where a is the first term, d is common difference, and n is the position of the term.

How do you identify if a list of numbers is an AP?

Check if the difference between any two consecutive terms is the same; if yes, it's an AP.

Can the common difference d be negative?

Yes, a negative d means the AP is decreasing (each term is smaller than the previous term).

What is a finite AP?

An AP that has a fixed number of terms and a last term (e.g., 45, 43, 41, ..., 31).

What is an infinite AP?

An AP that continues indefinitely without a last term (e.g., 1, 2, 3, 4, ...).

If a = 8 and d = 0, what kind of AP is formed?

A constant AP where all terms are the same: 8, 8, 8, 8, ... (common difference is zero).

What two pieces of information uniquely determine an AP?

The first term a and the common difference d; knowing both allows you to write any term or the entire sequence.

Important Board Questions

Write the first four terms of the AP where the first term a = 3 and the common difference d = -0.5. Is this AP finite or infinite? [2 marks]

Use the general form a, a+d, a+2d, a+3d to find the four terms. Then explain whether the AP stops at a specific term (finite) or continues forever (infinite) based on whether a last term is mentioned.

Reena's monthly salary starts at ₹8000 with an annual increment of ₹500. (a) How much will her salary be in the 5th year? (b) Write the AP representing her salary for the first 5 years. [3 marks]

Here a = 8000 and d = 500 (annual increment). Use the formula an = a + (n-1)d to find the 5th year salary. For part (b), list the terms: a, a+d, a+2d, a+3d, a+4d to show the AP for 5 years.

The ladder in Example (ii) has rungs whose lengths decrease uniformly by 2 cm from bottom to top, with the bottom rung being 45 cm. (a) Find the length of the 8th rung from the bottom. (b) Find the sum of the lengths of all 8 rungs. (c) Is this AP finite or infinite, and why? [5 marks]

Identify a = 45 (bottom rung), d = -2 (decrease), and n = 8. Use an = a + (n-1)d to find the 8th rung. For the sum, use Sn = n/2[2a + (n-1)d] or Sn = n/2(first + last). For part (c), explain that since the ladder has exactly 8 rungs, the AP has a fixed last term, making it finite.

Assertion-Reason Questions for Arithmetic Progressions

Q1. Assertion (A):

The sequence 5, 10, 15, 20, ... is an Arithmetic Progression.

Reason (R):

The difference between any two consecutive terms is constant and equal to 5.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both the assertion and reason are true, and the reason directly explains why the sequence is an AP. A constant common difference (d = 5) is the defining characteristic of an AP, making the reason the correct explanation for the assertion.

Q2. Assertion (A):

In the AP 20, 15, 10, 5, 0, ..., the common difference is -5.

Reason (R):

The sequence is decreasing, so the common difference must be negative.

Answer: B — Both A and R are true, but R is NOT the correct explanation of A — Both assertion and reason are true. The common difference is indeed d = 15 - 20 = -5. However, the reason (sequence is decreasing, therefore d is negative) is not the complete logical explanation. The common difference is -5 because we calculate a₂ - a₁ = 15 - 20 = -5, regardless of whether it's decreasing. The reason describes an observation rather than the defining principle.

Q3. Assertion (A):

The sequence 2, 4, 8, 16, ... is an Arithmetic Progression.

Reason (R):

Each term is obtained by adding a fixed number to the previous term.

Answer: D — A is false but R is true — The assertion is false because 2, 4, 8, 16, ... is a geometric progression where each term is multiplied by 2, not an AP. The reason is also false in this context because the sequence does not have a constant additive difference. The differences are 2, 4, 8, which are not constant.

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Case-Based Questions (CBQ) for Arithmetic Progressions

Case Study:

Reena got a job with a starting monthly salary of ₹8000. According to her contract, she receives an annual increment of ₹500. Her salary for the 1st year is ₹8000, 2nd year is ₹8500, 3rd year is ₹9000, and so on. Reena wants to calculate her salary progression over several years to plan her finances.

Q1. What is the common difference in Reena's salary progression?

  • A. ₹500 ✓
  • B. ₹800
  • C. ₹1000
  • D. ₹8000

Answer: A

Q2. What will be Reena's salary in the 6th year?

  • A. ₹10000
  • B. ₹10500 ✓
  • C. ₹11000
  • D. ₹11500

Answer: B

Q3. In which year will Reena's salary reach ₹12000?

  • A. 8th year
  • B. 9th year ✓
  • C. 10th year
  • D. 11th year

Answer: B

Q4. Identify the correct AP representation of Reena's salary scheme.

  • A. 8000, 8500, 9000, 9500, ... ✓
  • B. 500, 1000, 1500, 2000, ...
  • C. 8000, 9000, 10000, 11000, ...
  • D. 8000, 8000, 8000, 8000, ...

Answer: A

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Previous Year Questions (PYQ) for Arithmetic Progressions

Real CBSE board exam questions from past papers — free to attempt.

CBSE 2026 · Set 31/4/1 · 1 mark

The sum of first n terms of an A.P. is 502. If the first and the last terms are 2 and 192 respectively, then the value of n is:

  • 10
  • 5
  • 15
  • 20

CBSE 2026 · Set 31/4/1 · 1 mark

nth term of the A.P. –3/2, 3/2, 9/2, ... is:

  • 3n/2 – 3
  • 3n – 9/2
  • 3n – 9/2
  • 3n + 3/2

CBSE 2026 · Set 31/4/1 · 5 marks

The third and ninth term of an A.P. are 4 and – 8 respectively. (i) Which term of the A.P. is zero? (ii) Find the value of n if Sₙ = – 36.

CBSE 2026 · Set 31/4/2 · 1 mark

nth term of an A.P. is 5n – 15. The common difference of the A.P. is:

  • 5n
  • 5
  • – 5
  • 10

CBSE 2026 · Set 31/4/2 · 1 mark

nth term of the A.P. – 3/2, 3/2, 9/2, ... is:

  • 3n/2 – 3
  • 3n – 9/2
  • 3n – 9/2
  • 3n + 3/2

CBSE 2026 · Set 31/4/3 · 5 marks

The third and ninth term of an A.P. are 4 and –8 respectively. (i) Which term of the A.P. is zero? (ii) Find the value of n if Sn = –36.

CBSE 2026 · Set 31/5/1 · 1 mark

nth term of the A.P.: 4/3, –1/3, 3, ... is

  • 9/3 – 5n
  • 6/3 – 5n
  • 4/3 – 3n
  • 2/3 + 3n

CBSE 2026 · Set 31/5/1 · 1 mark

If –26, x, 2 are in A.P., then the value of x is

  • 14
  • –13
  • –12
  • –14

CBSE 2026 · Set 31/5/1 · 3 marks

In an A.P., it is given that a = 2, d = 8 and Sn = 90. Find the value of n.

CBSE 2026 · Set 31/5/2 · 1 mark

If 16/9, –2/9, –20/9, .... are in A.P., then next term of the sequence is

  • 32/9
  • 46/9
  • 2/9
  • 34/9

CBSE 2026 · Set 31/5/2 · 3 marks

(a) How many terms of the A.P. 3, 5, 7, 9, ... must be added to get the sum 80? OR (b) Find three consecutive terms in A.P. whose sum is 21 and their product is 231.

CBSE 2026 · Set 31/5/3 · 1 mark

If 14th term of an A.P. is 4 and its 15th term is zero, then its first term is

  • –48
  • –56
  • 56
  • 48

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Class 10 Maths Formula Sheet

Every formula from all 14 chapters on one page, checked against the NCERT textbook and the CBSE 2026-27 syllabus. Free PDF download.

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Class 10 Mathematics Previous Year Questions, Chapter-wise

All 1107 CBSE board questions from 2022–2026, sorted by chapter and question type. Free PDF download.

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