The common difference of the AP \(11, 7, 3, -1, \ldots\) is
- (a)\(4\)
- (b)\(-4\)
- (c)\(3\)
- (d)\(-3\)
Show answer
Answer: (b) \(-4\)
Why: d = a₂ − a₁ = 7 − 11.
\(d = 7 - 11 = -4\) (and \(3 - 7 = -4\), \(-1 - 3 = -4\)).
Arithmetic Progressions Class 10: MCQ and case study questions
15 multiple-choice questions and 2 case-based questions on the current (rationalised) syllabus, in the style of the board paper's Sections A and E. Try each one first; the answer, a one-line reason and a worked solution open on a tap.
- 15 MCQs (1 mark each)
- 2 case studies (4 marks each)
- No calculator needed
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Multiple-choice questions
Choose one option. Section A of the board paper has 18 MCQs of 1 mark each, spread over all chapters.
Which of the following lists forms an AP?
- (a)\(3, 6, 12, 24, \ldots\)
- (b)\(1, 8, 27, 64, \ldots\)
- (c)\(-5, -1, 3, 7, \ldots\)
- (d)\(2, \dfrac32, \dfrac43, \dfrac54, \ldots\)
Show answer
Answer: (c) \(-5, -1, 3, 7, \ldots\)
Why: Only −5, −1, 3, 7 has a constant difference (4).
Differences: \(3, 6, 12\); \(7, 19, 37\); \(4, 4, 4\); \(-\tfrac12, -\tfrac16, -\tfrac1{12}\). Only \(-5, -1, 3, 7\) has a constant difference.
The \(12\)th term of the AP \(5, 9, 13, \ldots\) is
- (a)\(49\)
- (b)\(45\)
- (c)\(53\)
- (d)\(48\)
Show answer
Answer: (a) \(49\)
Why: a₁₂ = a + 11d = 5 + 44.
\(a = 5\), \(d = 4\): \(a_{12} = 5 + 11 \times 4 = 49\).
The number of terms in the AP \(3, 8, 13, \ldots, 98\) is
- (a)\(19\)
- (b)\(20\)
- (c)\(21\)
- (d)\(25\)
Show answer
Answer: (b) \(20\)
Why: 98 = 3 + (n − 1)5 gives n − 1 = 19.
\(3 + (n - 1)5 = 98 \Rightarrow n - 1 = 19 \Rightarrow n = 20\).
If \(3k - 1\), \(2k + 4\) and \(4k\) are consecutive terms of an AP, then \(k\) equals
- (a)\(3\)
- (b)\(2\)
- (c)\(9\)
- (d)\(1\)
Show answer
Answer: (a) \(3\)
Why: The middle term is the average: 2(2k + 4) = (3k − 1) + 4k.
\(2(2k + 4) = 3k - 1 + 4k \Rightarrow 4k + 8 = 7k - 1 \Rightarrow k = 3\). Terms: \(8, 10, 12\).
The sum of the first \(15\) multiples of \(4\) is
- (a)\(240\)
- (b)\(520\)
- (c)\(600\)
- (d)\(480\)
Show answer
Answer: (d) \(480\)
Why: 4 + 8 + … + 60 = 15/2 × (4 + 60).
\(S_{15} = \dfrac{15}{2}(4 + 60) = \dfrac{15}{2} \times 64 = 480\).
If the \(n\)th term of an AP is \(a_n = 4n + 3\), then \(a_{10} - a_6\) equals
- (a)\(4\)
- (b)\(12\)
- (c)\(16\)
- (d)\(20\)
Show answer
Answer: (c) \(16\)
Why: a₁₀ − a₆ = 4d with d = 4.
\(a_{10} = 43\), \(a_6 = 27\); \(a_{10} - a_6 = 16 \ (= 4d)\).
The sum of the first \(n\) terms of an AP is \(S_n = 2n^2 + n\). Its \(8\)th term is
- (a)\(31\)
- (b)\(29\)
- (c)\(136\)
- (d)\(33\)
Show answer
Answer: (a) \(31\)
Why: a₈ = S₈ − S₇ = 136 − 105.
\(S_8 = 128 + 8 = 136\), \(S_7 = 98 + 7 = 105\). \(a_8 = 136 - 105 = 31\).
Which term of the AP \(3, 10, 17, \ldots\) is \(136\)?
- (a)\(19\)th
- (b)\(20\)th
- (c)\(21\)st
- (d)\(18\)th
Show answer
Answer: (b) \(20\)th
Why: 3 + 7(n − 1) = 136 gives n − 1 = 19.
\(3 + 7(n - 1) = 136 \Rightarrow 7(n - 1) = 133 \Rightarrow n = 20\).
The sum \(2 + 5 + 8 + \ldots + 32\) equals
- (a)\(170\)
- (b)\(204\)
- (c)\(176\)
- (d)\(187\)
Show answer
Answer: (d) \(187\)
Why: There are 11 terms, so S = 11/2 × (2 + 32).
\(32 = 2 + 3(n - 1) \Rightarrow n = 11\). \(S = \dfrac{11}{2}(2 + 32) = 11 \times 17 = 187\).
The \(3\)rd term of an AP is \(11\) and its \(8\)th term is \(31\). Its \(15\)th term is
- (a)\(55\)
- (b)\(59\)
- (c)\(63\)
- (d)\(60\)
Show answer
Answer: (b) \(59\)
Why: 5d = 31 − 11 gives d = 4, a = 3; a₁₅ = 3 + 56.
\(a_8 - a_3 = 5d = 20 \Rightarrow d = 4\); \(a = 11 - 2(4) = 3\). \(a_{15} = 3 + 14 \times 4 = 59\).
An AP has first term \(8\) and common difference \(-3\). Its first negative term is the
- (a)\(4\)th term
- (b)\(3\)rd term
- (c)\(5\)th term
- (d)\(9\)th term
Show answer
Answer: (a) \(4\)th term
Why: Terms: 8, 5, 2, −1, … so the 4th term is the first negative one.
\(a_n = 8 - 3(n - 1) \lt 0 \Rightarrow n - 1 \gt \dfrac83 \Rightarrow n \ge 4\). Indeed \(a_4 = -1\).
The middle term of the AP \(6, 11, 16, \ldots, 106\) is
- (a)\(51\)
- (b)\(61\)
- (c)\(56\)
- (d)\(53\)
Show answer
Answer: (c) \(56\)
Why: There are 21 terms, so the middle one is the 11th.
\(106 = 6 + 5(n - 1) \Rightarrow n = 21\). Middle term \(= a_{11} = 6 + 10 \times 5 = 56\).
How many terms of the AP \(1, 2, 3, \ldots\) must be added to get a sum of \(210\)?
- (a)\(21\)
- (b)\(19\)
- (c)\(15\)
- (d)\(20\)
Show answer
Answer: (d) \(20\)
Why: n(n + 1)/2 = 210 gives n² + n − 420 = 0 = (n + 21)(n − 20).
\(\dfrac{n(n + 1)}{2} = 210 \Rightarrow n^2 + n - 420 = 0 \Rightarrow (n + 21)(n - 20) = 0\), so \(n = 20\).
Three numbers in AP have sum \(27\) and product \(693\). The largest of them is
- (a)\(13\)
- (b)\(12\)
- (c)\(11\)
- (d)\(9\)
Show answer
Answer: (c) \(11\)
Why: Take a − d, a, a + d: 3a = 27, then 9(81 − d²) = 693.
\(3a = 27 \Rightarrow a = 9\). \(9(81 - d^2) = 693 \Rightarrow 81 - d^2 = 77 \Rightarrow d = \pm 2\). The numbers are \(7, 9, 11\); the largest is \(11\).
Case study questions
Section E of the board paper has three case-based questions of 4 marks: a real-life passage, then parts of 1, 1 and 2 marks, with a choice (OR) on the 2-mark part.
Case study 1: Weekly savings (4 marks)
Nisha starts saving for a new bicycle. She puts ₹\(50\) in her savings jar in the first week, ₹\(60\) in the second week, ₹\(70\) in the third week, and so on, adding ₹\(10\) more each week than the week before.
(i) How much does she put in the jar in the \(10\)th week? [1 mark]
Show answer
Answer: ₹\(140\)
\(a_{10} = 50 + 9 \times 10 = \text{₹}140\). A1
(ii) In which week does she put exactly ₹\(250\) in the jar? [1 mark]
Show answer
Answer: \(21\)st week
\(50 + 10(n - 1) = 250 \Rightarrow n = 21\). A1
(iii) Find the total amount she saves in the first \(20\) weeks. [2 marks]
Show answer
Answer: ₹\(2900\)
\(S_{20} = \dfrac{20}{2}[2(50) + 19(10)]\) M1 \(= 10 \times 290 = \text{₹}2900\). A1
OR After how many weeks will her total savings be exactly ₹\(1100\)? [2 marks]
Show answer
Answer: \(11\) weeks
\(\dfrac{n}{2}[100 + 10(n - 1)] = 1100 \Rightarrow n^2 + 9n - 220 = 0\) M1 \(\Rightarrow (n + 20)(n - 11) = 0\), so \(n = 11\). A1
Case study 2: Stadium stand (4 marks)
A new stand at a district cricket ground has \(25\) seats in the front row. Each row behind has \(3\) seats more than the row in front of it.
(i) How many seats are there in the \(12\)th row? [1 mark]
Show answer
Answer: \(58\)
\(a_{12} = 25 + 11 \times 3 = 58\). A1
(ii) Which row has \(85\) seats? [1 mark]
Show answer
Answer: \(21\)st row
\(25 + 3(n - 1) = 85 \Rightarrow n - 1 = 20 \Rightarrow n = 21\). A1
(iii) Find the total number of seats in the first \(20\) rows. [2 marks]
Show answer
Answer: \(1070\)
\(S_{20} = \dfrac{20}{2}[2(25) + 19(3)]\) M1 \(= 10 \times 107 = 1070\). A1
OR If the stand has \(30\) rows, how many seats does it have in all? [2 marks]
Show answer
Answer: \(2055\)
\(a_{30} = 25 + 29 \times 3 = 112\) M1; \(S_{30} = \dfrac{30}{2}(25 + 112) = 15 \times 137 = 2055\). A1
Next steps for Arithmetic Progressions
This free set is separate from the chapter's question bank. On the Arithmetic Progressions chapter page the revision notes and the first step of the Route to 95 are free for everyone. CBSE Essentials adds all 39 questions in the chapter bank (short and long answers, assertion–reason and more case studies) with their full step-marking schemes, the Route to 95 checkpoints with your progress saved, Skill Builders and the full common-mistakes library. There is no AI marking on CBSE Math Revision: you check your work against the marking scheme.
Original questions written by CBSE Math Revision for the CBSE 2026-27 syllabus; not taken from NCERT, NCERT Exemplar or CBSE papers. Every answer was re-checked by computer algebra and by an independent reviewer. CBSE Math Revision is independent and not affiliated with CBSE or NCERT. Spotted a slip? Tell us and it goes in the corrections log.