CBSE Class 10 · Chapter 5 · Algebra · 2026-27

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)
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Multiple-choice questions

Tap an option to check it; the answer and a worked solution open. Section A of the board paper has 18 MCQs of 1 mark each, spread over all chapters.

Q1

·1 mark·Multiple choiceIdentifying an AP

The common difference of the AP $11, 7, 3, - 1, \dots$ is

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Answer: (b)

- 4

Why: d = a₂ − a₁ = 7 − 11.

$d = 7 - 11 = - 4$ (and $3 - 7 = - 4$ , $- 1 - 3 = - 4$ ).

Q2

·1 mark·Multiple choiceIdentifying an AP

Which of the following lists forms an AP?

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Answer: (c)

- 5, - 1, 3, 7, \dots

Why: Only −5, −1, 3, 7 has a constant difference (4).

Differences: $3, 6, 12$ ; $7, 19, 37$ ; $4, 4, 4$ ; $- \frac{1}{2}, - \frac{1}{6}, - \frac{1}{12}$ . Only $- 5, - 1, 3, 7$ has a constant difference.

Q3

·1 mark·Multiple choicenth term of an AP

The $12$ th term of the AP $5, 9, 13, \dots$ is

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Answer: (a)

49

Why: a₁₂ = a + 11d = 5 + 44.

$a = 5$ , $d = 4$ : $a_{12} = 5 + 11 \times 4 = 49$ .

Q4

·1 mark·Multiple choicenth term of an AP

The number of terms in the AP $3, 8, 13, \dots, 98$ is

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Answer: (b)

20

Why: 98 = 3 + (n − 1)5 gives n − 1 = 19.

$3 + (n - 1) 5 = 98 \Rightarrow n - 1 = 19 \Rightarrow n = 20$ .

Q5

·1 mark·Multiple choiceIdentifying an AP

If $3 k - 1$ , $2 k + 4$ and $4 k$ are consecutive terms of an AP, then $k$ equals

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Answer: (a)

3

Why: The middle term is the average: 2(2k + 4) = (3k − 1) + 4k.

$2 (2 k + 4) = 3 k - 1 + 4 k \Rightarrow 4 k + 8 = 7 k - 1 \Rightarrow k = 3$ . Terms: $8, 10, 12$ .

Q6

·1 mark·Multiple choiceSum of first n terms

The sum of the first $15$ multiples of $4$ is

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Answer: (d)

480

Why: 4 + 8 + … + 60 = 15/2 × (4 + 60).

$S_{15} = \frac{15}{2} (4 + 60) = \frac{15}{2} \times 64 = 480$ .

Q7

·1 mark·Multiple choicenth term of an AP

If the $n$ th term of an AP is $a_{n} = 4 n + 3$ , then $a_{10} - a_{6}$ equals

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Answer: (c)

16

Why: a₁₀ − a₆ = 4d with d = 4.

$a_{10} = 43$ , $a_{6} = 27$ ; $a_{10} - a_{6} = 16 (= 4 d)$ .

Q8

·1 mark·Multiple choiceSum of first n terms

The sum of the first $n$ terms of an AP is $S_{n} = 2 n^{2} + n$ . Its $8$ th term is

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Answer: (a)

31

Why: a₈ = S₈ − S₇ = 136 − 105.

$S_{8} = 128 + 8 = 136$ , $S_{7} = 98 + 7 = 105$ . $a_{8} = 136 - 105 = 31$ .

Q9

·1 mark·Multiple choicenth term of an AP

Which term of the AP $3, 10, 17, \dots$ is $136$ ?

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Answer: (b)

20

Why: 3 + 7(n − 1) = 136 gives n − 1 = 19.

$3 + 7 (n - 1) = 136 \Rightarrow 7 (n - 1) = 133 \Rightarrow n = 20$ .

Q10

·1 mark·Multiple choiceSum of first n terms

The sum $2 + 5 + 8 + \dots + 32$ equals

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Answer: (d)

187

Why: There are 11 terms, so S = 11/2 × (2 + 32).

$32 = 2 + 3 (n - 1) \Rightarrow n = 11$ . $S = \frac{11}{2} (2 + 32) = 11 \times 17 = 187$ .

Q11

·1 mark·Multiple choicenth term of an AP

The $3$ rd term of an AP is $11$ and its $8$ th term is $31$ . Its $15$ th term is

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Answer: (b)

59

Why: 5d = 31 − 11 gives d = 4, a = 3; a₁₅ = 3 + 56.

$a_{8} - a_{3} = 5 d = 20 \Rightarrow d = 4$ ; $a = 11 - 2 (4) = 3$ . $a_{15} = 3 + 14 \times 4 = 59$ .

Q12

·1 mark·Multiple choicenth term of an AP

An AP has first term $8$ and common difference $- 3$ . Its first negative term is the

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Answer: (a)

4

th term

Why: Terms: 8, 5, 2, −1, … so the 4th term is the first negative one.

$a_{n} = 8 - 3 (n - 1) < 0 \Rightarrow n - 1 > \frac{8}{3} \Rightarrow n \geq 4$ . Indeed $a_{4} = - 1$ .

Q13

·1 mark·Multiple choicenth term of an AP

The middle term of the AP $6, 11, 16, \dots, 106$ is

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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 $= a_{11} = 6 + 10 \times 5 = 56$ .

Q14

·1 mark·Multiple choiceSum of first n terms

How many terms of the AP $1, 2, 3, \dots$ must be added to get a sum of $210$ ?

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Answer: (d)

20

Why: n(n + 1)/2 = 210 gives n² + n − 420 = 0 = (n + 21)(n − 20).

$\frac{n (n + 1)}{2} = 210 \Rightarrow n^{2} + n - 420 = 0 \Rightarrow (n + 21) (n - 20) = 0$ , so $n = 20$ .

Q15

·1 mark·Multiple choiceIdentifying an AP

Three numbers in AP have sum $27$ and product $693$ . The largest of them is

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Answer: (c)

11

Why: Take a − d, a, a + d: 3a = 27, then 9(81 − d²) = 693.

$3 a = 27 \Rightarrow a = 9$ . $9 (81 - d^{2}) = 693 \Rightarrow 81 - d^{2} = 77 \Rightarrow d = \pm 2$ . 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. All Class 10 case study questions by chapter.

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]

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Answer: ₹

140

a_{10} = 50 + 9 \times 10 = ₹ 140

. A1

(ii) In which week does she put exactly ₹

250

in the jar? [1 mark]

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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]

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Answer: ₹

2900

S_{20} = \frac{20}{2} [2 (50) + 19 (10)]

= 10 \times 290 = ₹ 2900

. A1

OR After how many weeks will her total savings be exactly ₹

1100

? [2 marks]

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Answer:

11

weeks

\frac{n}{2} [100 + 10 (n - 1)] = 1100 \Rightarrow n^{2} + 9 n - 220 = 0

\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]

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Answer:

58

a_{12} = 25 + 11 \times 3 = 58

. A1

(ii) Which row has

85

seats? [1 mark]

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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]

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Answer:

1070

S_{20} = \frac{20}{2} [2 (25) + 19 (3)]

= 10 \times 107 = 1070

. A1

OR If the stand has

30

rows, how many seats does it have in all? [2 marks]

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Answer:

2055

a_{30} = 25 + 29 \times 3 = 112

M1;

S_{30} = \frac{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 Pro adds all 44 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, the full common-mistakes library and AI marking of your written working.

Arithmetic Progressions: important questions What Pro includes

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 checked twice, the second time 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.