Arithmetic Progressions — Class 10 Maths MCQ with Answers (2026-27)

Updated 23 August 2026 · checked against the CBSE 2026-27 syllabus

16 objective questions on Arithmetic Progressions, every one checked against the CBSE 2026-27 syllabus. Pick an option and the correct answer is shown straight away, with your marks running at the top. A new question opens each day until all 16 are unlocked.

Arithmetic Progressions MCQ questions with answers

1. Veer can currently run 200 m in 51 seconds and with each day of practice it takes him 2 seconds less. He wants to do it in 31 seconds. Which of the following terms are in AP for​ the given situation?‌
   1. 51, 53, 55…
   2. 51, 49, 47…
   3. −51, −53, −55…
   4. 51, 55, 59…

   Answer: b) 51, 49, 47…

   He starts at 51 seconds and each day takes 2 seconds less, so the times decrease: 51, 49, 47, … with common difference −2.

2. What is the minimum number of‌ days he needs‌ to practise till his goal is achieved?
   1. 10
   2. 12
   3. 11
   4. 9

   Answer: c) 11

   The times form 51, 49, 47, …, 31. Counting the terms of this AP: (51 − 31)/2 + 1 = 11.

3. Which of the following terms is not in the AP of the‌ above​ given situation?
   1. 41
   2. 30
   3. 37
   4. 39

   Answer: b) 30

   Starting at 51 and subtracting 2 each time keeps every term odd. 30 is even, so it can never appear.

4. If the nth term of an AP is given by aₙ = 2n + 3,​ then‌ the common difference of the AP is
   1. 2
   2. 3
   3. 5
   4. 1

   Answer: a) 2

   The coefficient of n in aₙ = 2n + 3 is the common difference. Check: a₂ − a₁ = 7 − 5 = 2.

5. The value of x for which‌ 2x,‌ x + 10, 3x + 2 are three consecutive terms of an AP is
   1. 6
   2. −6
   3. 18
   4. −18

   Answer: a) 6

   In an AP consecutive differences are equal: (x + 10) − 2x = (3x + 2) − (x + 10). This gives 10 − x = 2x − 8, so x = 6.

6. Your elder brother repays a total loan of Rs 1,18,000 by paying every month, starting with a first instalment of Rs 1000 and increasing the instalment by​ Rs​ 100 every month. The amount paid by him in the 30th instalment is
   1. 3900
   2. 3500
   3. 3700
   4. 3600

   Answer: a) 3900

   a = 1000, d = 100, so a₃₀ = 1000 + 29(100) = Rs 3900.

7. The amount paid by him in the 30‌ instalments​ is
   1. 37000
   2. 73500
   3. 75300
   4. 75000

   Answer: b) 73500

   S₃₀ = 30/2[2(1000) + 29(100)] = 15(4900) = Rs 73500.

8. What amount does he still​ have‌ to pay after the 30th instalment?
   1. 45500
   2. 49000
   3. 44500
   4. 54000

   Answer: c) 44500

   He has paid Rs 73500 of Rs 118000, leaving 118000 − 73500 = Rs 44500.

9. If the total instalments are 40, then what is the‌ amount‌ paid in the last instalment?
   1. 4900
   2. 3900
   3. 5900
   4. 9400

   Answer: a) 4900

   a₄₀ = 1000 + 39(100) = Rs 4900.

10. The ratio of the 1st instalment​ to​ the last instalment is
    1. 1:49
    2. 10:49
    3. 10:39
    4. 39:10

    Answer: b) 10:49

    First instalment 1000, last 4900. The ratio 1000 : 4900 simplifies to 10 : 49.

11. The 6th term of the AP −11, −8,​ −5,‌ … is
    1. −7
    2. 4
    3. 7
    4. 16

    Answer: b) 4

    a = −11 and d = 3, so a₆ = −11 + 5(3) = 4.

12. If the common difference of an AP​ is​ 7, then find the value of a₇ − a₄.
    1. 7
    2. 14
    3. 21
    4. 24

    Answer: c) 21

    a₇ − a₄ spans three common differences, so it equals 3d = 3(7) = 21. The first term never matters.

13. The next term‌ of‌ the AP √3, √12, √27, … is
    1. √9
    2. √15
    3. √48
    4. √12

    Answer: c) √48

    Simplify first: √3, 2√3, 3√3 — an AP with d = √3. The next term is 4√3 = √48.

14. A safe seating-standing section of a stadium has 20 rows. Each row has one more seat than the previous row, starting from the second row. The first row has 4 seats.​ Sidharth​ is seating in the centre seat of Row 12. How many seats are on his left?
    1. 5
    2. 7
    3. 8
    4. 24

    Answer: b) 7

    Row 12 has 4 + 11 = 15 seats. Removing the centre seat leaves 14, split equally, so 7 are on his left.

15. What is the seating‌ capacity​ of the section?
    1. 80
    2. 210
    3. 270
    4. 840

    Answer: c) 270

    Seats form the AP 4, 5, 6, … for 20 rows. S₂₀ = 20/2[2(4) + 19(1)] = 10(27) = 270.

16. What cannot be the difference between four consecutive terms of​ an​ arithmetic progression?
    1. 0, 0, 0
    2. −2, −2, −2
    3. 2, 3, 4
    4. 2/7, 2/7, 2/7

    Answer: c) 2, 3, 4

    In an AP every consecutive difference is the same. 2, 3, 4 are not equal, so those terms cannot be in AP. Zero and fractions are perfectly valid common differences.