Arithmetic Progression CBQs — CBSE Class 10 (2027) with step-wise solutions

Sample Arithmetic Progression CBQs

1. Question 1MCQ1 mark
   In a community hall the first row has 24 chairs and each row behind it has 2 chairs more than the row in front of it. Which row has exactly 60 chairs?
   • 17th row
   • 18th row
   • 19th row
   • 20th row
   Show step-wise solution

   1. [1 mark] Chairs per row form an AP with a = 24, d = 2; 24 + (n − 1) × 2 = 60 ⇒ n − 1 = 18 ⇒ n = 19, so it is the 19th row.

   Answer: 19th row

2. Question 2Assertion-Reason1 mark
   A school auditorium has 30 rows. The 1st row has 20 seats and each row has 2 seats more than the row in front of it. Assertion (A): The 30th row has 78 seats. Reason (R): The sum of the first n terms of an AP is Sₙ = n/2 [2a + (n − 1)d].
   • (A) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
   • (B) Both Assertion and Reason are true but Reason is NOT the correct explanation of Assertion.
   • (C) Assertion is true but Reason is false.
   • (D) Assertion is false but Reason is true.
   Show step-wise solution

   1. [1 mark] (B): a₃₀ = a + 29d = 20 + 29 × 2 = 78, so A is true; R is a true formula, but A uses the nth-term formula aₙ = a + (n − 1)d, not the sum formula, so R does not explain A.

   Answer: (B)

3. Question 3Short answer2 marks
   A new metro line carried 4200 passengers on its first day of operation. Its ridership grew steadily, with 150 more passengers each day than on the previous day. The operator will add an extra train from the first day on which the ridership reaches 6000. On which day will the extra train be added?
   Show step-wise solution

   1. [1 mark] Daily ridership is an A.P. with a = 4200 and d = 150, so the ridership on day n is aₙ = 4200 + (n − 1) × 150.
   2. [1 mark] 4200 + (n − 1) × 150 = 6000 ⇒ (n − 1) × 150 = 1800 ⇒ n − 1 = 12 ⇒ n = 13. The extra train is added on day 13.

   Answer: Day 13

4. Question 4Case study4 marks
   A stepwell (baori) in a Rajasthan village is being restored. Its front wall has 15 horizontal stone steps going down to the water. The topmost step is 42 m long and each step below it is 2 m shorter than the step just above it. The restoration team will fix a brass strip along the front edge of every step. Based on the above information, answer the following questions: (i) Find the length of the 8th step from the top. [1 mark] (ii) Which step from the top is 18 m long? [1 mark] (iii) Find the total length of brass strip needed for all 15 steps. [2 marks]
   Show step-wise solution

   1. [1 mark] (i) The lengths form an AP with a = 42 m and d = −2 m; a₈ = 42 + 7 × (−2) = 28 m.
   2. [1 mark] (ii) 42 + (n − 1)(−2) = 18 ⇒ n − 1 = 12 ⇒ n = 13, so the 13th step is 18 m long.
   3. [1 mark] (iii) S₁₅ = (15/2)[2 × 42 + 14 × (−2)] = (15/2)(84 − 28) = (15/2) × 56.
   4. [1 mark] (iii) S₁₅ = 420, so 420 m of brass strip is needed.

   Answer: (i) 28 m (ii) 13th step (iii) 420 m

5. Question 5Long answer5 marks
   Riya is preparing for a district half-marathon. Her coach sets a plan in which she runs 1.5 km on Day 1 and, on each following day, runs 0.25 km more than on the previous day. (a) How far does she run on Day 15? (b) On which day does she run exactly 6 km? (c) Find the total distance she runs in the first 20 days.
   Show step-wise solution

   1. [1 mark] The daily distances 1.5, 1.75, 2, … km form an A.P. with a = 1.5 and d = 0.25, so aₙ = 1.5 + (n − 1)(0.25).
   2. [1 mark] (a) a₁₅ = 1.5 + 14 × 0.25 = 1.5 + 3.5 = 5 km.
   3. [1 mark] (b) 1.5 + (n − 1)(0.25) = 6 ⇒ (n − 1)(0.25) = 4.5 ⇒ n − 1 = 18 ⇒ n = 19, so she runs 6 km on Day 19.
   4. [1 mark] (c) S₂₀ = 20/2 [2(1.5) + 19(0.25)] = 10[3 + 4.75].
   5. [1 mark] S₂₀ = 10 × 7.75 = 77.5 km.

   Answer: (a) 5 km (b) Day 19 (c) 77.5 km

Study Arithmetic Progression

• Arithmetic Progression notes
• Arithmetic Progression Topic Hub
• All Class 10 CBQs