Patterns and sequences are everywhere in mathematics, and Arithmetic Progressions is one of the first chapters where students start understanding how numbers follow a fixed logical pattern. This chapter is considered highly important in Class 10 Maths because it teaches students how to identify relationships between numbers, calculate terms using formulas, and solve sequence-based problems accurately. Practicing Arithmetic Progressions MCQs helps students improve calculation speed, pattern recognition, and formula application skills for CBSE Board exams. The latest exam pattern now focuses more on competency-based learning and logical understanding, which makes regular objective question practice extremely important for improving accuracy and confidence. Students preparing for board examinations can also explore Class 10 MCQs for chapter-wise objective practice based on the latest CBSE pattern and conceptual learning approach.
Why Arithmetic Progressions is an Important Chapter in Class 10 Maths
Arithmetic Progressions is one of the most scoring chapters in Class 10 Mathematics because the formulas are direct, concepts are structured, and logical number patterns make calculations easier once students understand the basics properly.
This chapter is important because:
- Questions are frequently asked in CBSE board exams
- Formula-based calculations improve accuracy
- Students learn pattern-recognition skills
- Sequence understanding becomes stronger
- Logical mathematical thinking improves
- Competency-based questions are increasing
- Numerical reasoning skills become better
Students who understand arithmetic progressions properly usually solve sequence-related questions much faster in examinations.
Main Concepts Covered in Arithmetic Progressions
Before solving Arithmetic Progressions MCQs with Answers, students should revise all important concepts carefully because many objective questions are directly formula and pattern based.
Important topics covered in this chapter include:
- Arithmetic progression basics
- Meaning of sequence
- First term of AP
- Common difference
- nth term formula
- Sum of n terms
- Increasing arithmetic progression
- Decreasing arithmetic progression
- Finite sequence
- Consecutive terms
- Pattern observation
- Formula-based calculations
- Sequence simplification
A proper understanding of these concepts helps students solve board-level MCQs more confidently.
Arithmetic Progressions MCQs with Answers
Practice important and exam-oriented Arithmetic Progressions MCQs designed according to the latest CBSE pattern and competency-based learning approach. These objective questions help students improve formula application, sequence analysis, numerical reasoning, and calculation accuracy for school exams and board preparation.
Q. In an Arithmetic Progression, the common difference is -4 and the 7th term is 4. Find the first term.
A) 24
B) 20
C) 28
D) 32
Answer: C
Explanation: Using the formula:
an = a + (n - 1)d
4 = a + (7 - 1)(-4)
4 = a - 24
a = 28
Q. If the first term of an AP is 3.5 and the common difference is 0, then the 101st term is:
A) 0
B) 3.5
C) 103.5
D) 101
Answer: B
Explanation: When d = 0, all terms remain equal.
a101 = 3.5 + (101 - 1)(0)
a101 = 3.5
Q. The sequence -10, -6, -2, 2, ... is:
A) An AP with d = -4
B) Not an AP
C) An AP with d = 4
D) A geometric progression
Answer: C
Explanation:
-6 - (-10) = 4
-2 - (-6) = 4
2 - (-2) = 4
Since the difference is constant, it is an AP.
Q. Find the 11th term of the AP: -5, -5/2, 0, 5/2, ...
A) -20
B) 20
C) 30
D) -30
Answer: B
Explanation:
a = -5
d = -5/2 - (-5)
d = -5/2 + 10/2
d = 5/2
a11 = a + 10d
a11 = -5 + 10(5/2)
a11 = -5 + 25
a11 = 20
Q. Which set correctly represents the first four terms of an AP having first term -2 and common difference -2?
A) -2, 0, 2, 4
B) -2, -4, -6, -8
C) -2, -6, -10, -14
D) -2, -4, -8, -16
Answer: B
Explanation:
a1 = -2
a2 = -2 + (-2) = -4
a3 = -4 + (-2) = -6
a4 = -6 + (-2) = -8
Q. The first two terms of an AP are -3 and 4. Find the 21st term.
A) 137
B) 143
C) 17
D) -143
Answer: A
Explanation:
a = -3
d = 4 - (-3)
d = 7
a21 = a + 20d
a21 = -3 + 20(7)
a21 = -3 + 140
a21 = 137
Q. In an AP, the 2nd term is 13 and the 5th term is 25. Find the 7th term.
A) 30
B) 33
C) 37
D) 38
Answer: B
Explanation:
a + d = 13
a + 4d = 25
Subtracting:
3d = 12
d = 4
a + 4 = 13
a = 9
a7 = a + 6d
a7 = 9 + 24
a7 = 33
Q. Which term of the AP 21, 42, 63, 84, ... is 210?
A) 9th
B) 10th
C) 11th
D) 12th
Answer: B
Explanation:
a = 21
d = 21
an = a + (n - 1)d
210 = 21 + (n - 1)(21)
210 = 21n
n = 10
Q. If the common difference of an AP is 5, then find: a18 - a13
A) 5
B) 20
C) 25
D) 30
Answer: C
Explanation:
a18 = a + 17d
a13 = a + 12d
a18 - a13 = 5d
= 5 × 5
= 25
Q. The nth term of an AP is given by: an = 3 + 4n. Find the common difference.
A) 7
B) 3
C) 4
D) 1
Answer: C
Explanation:
a1 = 3 + 4(1)
a1 = 7
a2 = 3 + 4(2)
a2 = 11
d = 11 - 7
d = 4
Q. If the first three terms of an AP are: 3y - 1, 3y + 5, 5y + 1
then find y.
A) 2
B) 3
C) 4
D) 5
Answer: D
Explanation:
(3y + 5) - (3y - 1) = (5y + 1) - (3y + 5)
6 = 2y - 4
2y = 10
y = 5
Q. Find the common difference of: 1/2q, (1 - 2q)/2q, (1 - 4q)/2q, ...
A) -1
B) 1
C) q
D) -q
Answer: A
Explanation:
d = (1 - 2q)/2q - 1/2q
d = (1 - 2q - 1)/2q
d = -2q/2q
d = -1
Q. If 18, a, b, -3 are in AP, then find a and b.
A) 11, 4
B) 12, 6
C) 13, 8
D) 14, 7
Answer: A
Explanation:
-3 = 18 + 3d
3d = -21
d = -7
a = 18 - 7
a = 11
b = 11 - 7
b = 4
Q. Find the 4th term from the end of the AP: -11, -8, -5, ..., 49
A) 37
B) 40
C) 43
D) 46
Answer: B
Explanation:
Last term = 49
d = 3
Required term = 49 - (4 - 1)(3)
= 49 - 9
= 40
Q. If the first term of an AP is 3 and common difference is 4, find the 10th term.
A) 39
B) 43
C) 47
D) 53
Answer: A
Explanation:
a10 = 3 + (10 - 1)(4)
= 3 + 36
= 39
Q. The sum of the first 15 terms of an AP is 255. If the first term is 3, find the common difference.
A) 1
B) 2
C) 3
D) 4
Answer: B
Explanation:
Using:
Sn = n/2 [2a + (n - 1)d]
255 = 15/2 [2(3) + 14d]
255 = 15/2 [6 + 14d]
34 = 6 + 14d
14d = 28
d = 2
Q. The sum of the first n terms of an AP is: Sn = 6n² - 2n. Find n when Sn = 1320.
A) 12
B) 13
C) 14
D) 15
Answer: D
Explanation:
6n² - 2n = 1320
3n² - n - 660 = 0
3n² - 45n + 44n - 660 = 0
3n(n - 15) + 44(n - 15) = 0
(n - 15)(3n + 44) = 0
n = 15
Q. Determine the first term and common difference of the AP: 3, 1, -1, -3, ...
A) 1, 3
B) -1, 3
C) 3, -2
D) 2, 3
Answer: C
Explanation:
First term:
a = 3
Common difference:
d = 1 - 3
d = -2
Q. If the (p + q)th term of an AP is m and the (p - q)th term is n, then the pth term equals:
A) m + n
B) (m - n)/2
C) (m + n)/2
D) mn
Answer: C
Explanation:
ap+q = a + (p + q - 1)d = m
ap-q = a + (p - q - 1)d = n
Adding both:
2a + (2p - 2)d = m + n
2[a + (p - 1)d] = m + n
ap = (m + n)/2
Q. Find the sum of all two-digit odd numbers.
A) 2425
B) 2475
C) 2525
D) 2575
Answer: B
Explanation:
AP:
11, 13, 15, ..., 99
a = 11
l = 99
d = 2
99 = 11 + (n - 1)2
88 = 2(n - 1)
n = 45
Sum:
S = n/2 (a + l)
S = 45/2 (11 + 99)
S = 45/2 × 110
S = 2475
Q. The sum of the first n odd natural numbers is:
A) 2n²
B) 2n + 1
C) 2n - 1
D) n²
Answer: D
Explanation:
1 + 3 + 5 + ... + (2n - 1)
Sn = n/2 [2 + (n - 1)2]
= n/2 (2n)
= n²
Q. If a, b, and c are in AP, then which relation is correct?
A) a + c = 2b
B) a + b = 2c
C) c = (a + b)/2
D) a + c = b
Answer: A
Explanation:
For AP:
b - a = c - b
2b = a + c
a + c = 2b
Q. The sum of the 6th and 7th terms of an AP is 39. If d = 3, find the first term.
A) 2
B) -3
C) 4
D) 3
Answer: D
Explanation:
a6 = a + 5d
a7 = a + 6d
(a + 5d) + (a + 6d) = 39
2a + 11d = 39
2a + 33 = 39
2a = 6
a = 3
Q. Three numbers are in AP and their sum is 30. If the greatest number is 13, find the common difference.
A) 2
B) 4
C) 5
D) 3
Answer: D
Explanation:
Let numbers be:
a - d, a, a + d
Sum:
3a = 30
a = 10
Largest number:
10 + d = 13
d = 3
Q. Find the common difference of the AP: 1/8, 2/8, 3/8, ...
A) 1/8
B) 1
C) 1/4
D) 1/2
Answer: A
Explanation:
d = 2/8 - 1/8
d = 1/8
Q. How many natural numbers less than or equal to 300 are divisible by 17?
A) 13
B) 15
C) 17
D) 19
Answer: C
Explanation:
Multiples of 17:
17, 34, 51, ..., 289
289 = 17 × 17
Hence, total numbers = 17
Q. Find the first negative term of the AP: 81/5, 77/5, 73/5, ...
A) 20th
B) 21st
C) 22nd
D) 23rd
Answer: C
Explanation:
a = 81/5
d = -4/5
an = a + (n - 1)d
81/5 + (n - 1)(-4/5) < 0
81 - 4n + 4 < 0
85 < 4n
n > 21.25
First integer value:
n = 22
Q. If: Sn = n(n - 1) then the nth term of the AP is:
A) 2n
B) 2n - 1
C) 2n - 2
D) 2n - 4
Answer: C
Explanation:
an = Sn - Sn-1
an = n(n - 1) - (n - 1)(n - 2)
an = (n² - n) - (n² - 3n + 2)
an = 2n - 2
Q. The first term of an AP is -12 and the 6th term is 8. If the sum of first n terms is 120, find n.
A) 10
B) 11
C) 12
D) 13
Answer: C
Explanation:
a = -12
a6 = a + 5d
8 = -12 + 5d
5d = 20
d = 4
Now:
120 = n/2 [2(-12) + (n - 1)4]
240 = n(-24 + 4n - 4)
240 = n(4n - 28)
4n² - 28n - 240 = 0
n² - 7n - 60 = 0
(n - 12)(n + 5) = 0
n = 12
Q. If 2x, x + 10, and 3x + 2 are consecutive terms of an AP, then x equals:
A) 4
B) 5
C) 6
D) 8
Answer: C
Explanation: For consecutive terms in AP:
2(x + 10) = 2x + (3x + 2)
2x + 20 = 5x + 2
18 = 3x
x = 6
Understanding Arithmetic Progressions in Simple Language
An Arithmetic Progression, also called AP, is a sequence of numbers in which the difference between consecutive terms always remains constant.
For example:
2, 4, 6, 8, 10
15, 12, 9, 6
3, 7, 11, 15
In these sequences, the difference between every two consecutive terms is fixed. This fixed value is called the common difference.
Examples:
In 2, 4, 6, 8 → common difference = 2
In 15, 12, 9, 6 → common difference = -3
Arithmetic Progressions help students understand how number patterns behave mathematically and how formulas can be used to calculate terms without writing the entire sequence manually.
Important Formulas Used in Arithmetic Progressions
nth Term Formula
an = a + (n − 1)d
This formula is used to calculate any specific term of the arithmetic progression directly.
Sum of n Terms Formula
Sn = n/2 [2a + (n − 1)d]
This formula helps calculate the sum of multiple terms quickly without adding every term individually.
Meaning of Formula Variables
a = first term
d = common difference
n = number of terms
an = nth term
Sn = sum of n terms
Students should practice formulas regularly because formula-based objective questions are common in CBSE board exams.
Instructions for Solving Arithmetic Progressions MCQs
- Read the sequence carefully before solving because many students make mistakes while identifying the pattern incorrectly.
- Check the common difference properly between consecutive terms before applying formulas.
- Identify whether the progression is increasing or decreasing because the sign of the common difference matters.
- Use formulas carefully and substitute values step-by-step to avoid arithmetic mistakes.
- Count terms properly while solving nth term and sum-related questions because incorrect term counting changes the final answer completely.
- Verify calculations wherever possible to improve accuracy in objective questions.
- Avoid rushing calculations because most mistakes happen due to careless simplification.
Practice competency-based and formula-application questions regularly because the latest CBSE pattern focuses heavily on conceptual understanding.
Mistakes Students Make in Arithmetic Progressions MCQs
Many students lose marks in Arithmetic Progressions MCQs because of small calculation or observation errors.
Some common mistakes include:
- Wrong common difference calculation
- Incorrect term counting
- Formula substitution mistakes
- Sign errors in decreasing AP
- Confusion between nth term and sum formulas
- Arithmetic simplification errors
- Skipping calculation steps
Students should solve questions carefully instead of depending only on mental calculations.
How to Identify Patterns in Arithmetic Progressions Quickly
One of the most important skills in this chapter is recognizing numerical patterns quickly.
Students can improve pattern identification by:
- Observing the gap between consecutive numbers
- Checking whether the difference remains constant
- Identifying increasing or decreasing trends
- Looking for repeated addition or subtraction patterns
- Comparing neighboring terms carefully
Once the common difference is identified correctly, most AP questions become much easier to solve.
Why Students Find Arithmetic Progressions Difficult
Although Arithmetic Progressions looks formula-based initially, many students still find the chapter confusing because of calculation pressure and sequence observation mistakes.
Common reasons include:
- Formula confusion
- Incorrect common difference
- Careless arithmetic errors
- Wrong nth term calculations
- Difficulty in large-term calculations
- Confusion between formulas
However, regular practice and proper understanding of number patterns make the chapter much easier over time.
Revision Notes for Arithmetic Progressions
Students should revise these important points regularly before exams:
- Arithmetic progression has constant common difference
- nth term formula: an = a + (n − 1)d
- Sum formula: Sn = n/2 [2a + (n − 1)d]
- Positive difference → increasing AP
- Negative difference → decreasing AP
- Careful term counting is important
- Formula substitution requires accuracy
- Sequence observation improves solving speed
Short revision sessions improve retention and board exam confidence significantly.
Conclusion
Practicing Arithmetic Progressions MCQs with Answers is one of the best ways to improve sequence understanding, formula application, and numerical reasoning skills for Class 10 Maths. This chapter helps students develop logical thinking by teaching how number patterns behave mathematically and how formulas simplify lengthy calculations. Students preparing for CBSE board exams should focus on conceptual clarity, regular objective practice, and careful formula application instead of memorizing steps blindly. Consistent practice improves calculation accuracy, confidence, speed, and overall board exam performance naturally.