Case Study Questions on Arithmetic Progression Chapter 5 for Class 10 with Answers

This set of case-study questions for Chapter 5: Arithmetic Progression (Class 10) follows CBSE and NCERT guidelines and presents short, exam-style scenarios designed to strengthen reasoning, pattern recognition, and stepwise problem solving. Problems cover identification of APs, deriving and using the nth-term and sum formulas, solving contextual word problems, handling mixed sequences, and translating real-life situations into AP models. A downloadable PDF accompanies the pack with practice sets for quick revision, classroom worksheets for practice focused on building speed and accuracy.

Solved Arithmetic Progression Case Study Questions and Answers

CBSE's case-based questions usually give you a short passage describing a real situation, sometimes with a small figure, followed by 4 - 5 questions of increasing difficulty, a couple of MCQs to check basic understanding, a short-answer question that needs a calculation, and a longer question that asks you to interpret or justify your answer.

Case Study 1:The Auditorium Seating Plan

A school is designing a new auditorium for its annual function. The first row, closest to the stage, has 18 seats. Every row behind it has 4 more seats than the row in front, because the hall widens as it goes back. There are 20 rows in total.

1. What is the common difference of this AP?

(a) 18

(b) 4

(c) 20

(d) 22

2. The number of seats in the 12th row is __________.

3. True / False

The total seating capacity of the auditorium (all 20 rows) is 1,120 seats.

4. Which row is the first to have more than 70 seats?

5. The school later decides to add 2 extra rows at the back, continuing the same pattern. Find the number of seats in the new last row (row 22).

Solutions:

1. Each row gains a fixed 4 seats over the previous row, so d = 4. Answer: (b) 4

2.  Using aₙ = a + (n−1)d with a = 18, d = 4, n = 12:

a₁₂ = 18 + (12−1)(4) = 18 + 44 = 62

Answer: 62 seats

3. S₂₀ = n/2 [2a + (n−1)d] = 20/2 [2(18) + 19(4)]

 = 10 [36 + 76] = 10 × 112 = 1,120

Answer: True

4.  We need aₙ > 70: 18 + (n−1)(4) > 70 

⇒(n−1)(4) > 52 

⇒ n−1 > 13 

⇒ n > 14

The smallest whole number satisfying this is n = 15. Checking: a₁₅ = 18 + 14(4) = 18 + 56 = 74, which is indeed over 70, while a₁₄ = 70 exactly (not "more than" 70).

Answer: Row 15

5.  a₂₂ = 18 + (22−1)(4) = 18 + 84 = 102

Answer: 102 seats

Case Study 2: The Pipe Stacking Yard

At a construction yard, cylindrical pipes are stacked in a triangular pile against a wall. The bottom row has 24 pipes. Each row above has 2 fewer pipes than the row below it, since the pile narrows as it rises. The stacking continues until a single pipe sits at the top.

1. What type of sequence do the number of pipes in each row form?

(a) Geometric Progression

(b) Arithmetic Progression with negative common difference

(c) Arithmetic Progression with positive common difference

(d) Not a defined sequence

2. The total number of rows in the pile is __________.

3. True / False: The total number of pipes used in the entire pile is 156.

4. How many pipes are in the 6th row from the bottom?

5. If the yard manager wants exactly 100 pipes used in the pile (counting from the bottom row upward), after how many rows should the stacking stop?

Solutions: 

The count decreases by a fixed 2 pipes per row, which is still an AP just with d = −2.

Answer: (b) Arithmetic Progression with negative common difference

2.  We need aₙ = 1: 24 + (n−1)(−2) = 1 

⇒ −2(n−1) = −23 

⇒ n−1 = 11.5

Since rows must reduce by exactly 2 each time starting at 24 (an even number), the pile actually reaches 2 pipes at the top, not 1, because 24 is even and subtracting 2 repeatedly always keeps the count even. So the last row has 2 pipes: 

24 + (n−1)(−2) = 2 

⇒ n−1 = 11 

⇒ n = 12.

Answer: 12 rows (ending at 2 pipes on top, not 1, since the count stays even)

3. S₁₂ = n/2 (a + l) = 12/2 (24 + 2) = 6 × 26 = 156

Answer: True

4. a₆ = 24 + (6−1)(−2) = 24 − 10 = 14

Answer: 14 pipes

5. Sₙ = n/2 [2(24) + (n−1)(−2)] = 100 

⇒ n/2 [48 − 2n + 2] = 100 

⇒ n(50 − 2n) = 200 

⇒ 50n − 2n² = 200

2n² − 50n + 200 = 0 

⇒ n² − 25n + 100 = 0 

⇒ (n − 5)(n − 20) = 0 

⇒ n = 5 or n = 20

Since the pile only has 12 rows in total, n = 20 is rejected.

Answer: After 5 rows

Case Study 3: The Marathon Training Plan

Rohan is training for his first half-marathon. In Week 1, he runs 5 km. To build endurance gradually, he increases his weekly running distance by a fixed amount of 1.5 km every week, for 14 weeks leading up to race day.

1. How far will Rohan run in Week 10?

(a) 16.5 km

(b) 18.5 km

(c) 19.5 km

(d) 15 km

2. The total distance Rohan runs across all 14 weeks is __________ km.

3. True / False

Rohan's running distance will cross 20 km in a single week before his training plan ends.

4. In which week does Rohan first run more than 20 km?

5. Rohan's coach suggests a tougher plan where he starts at the same 5 km but increases by 2 km every week instead. How much more would he run in total over the same 14 weeks under this tougher plan?

Solutions:

a₁₀ = 5 + (10−1)(1.5) = 5 + 13.5 = 18.5 km

Answer: (b) 18.5 km

2. S₁₄ = 14/2 [2(5) + 13(1.5)] = 7 [10 + 19.5] = 7 × 29.5 = 206.5

Answer: 206.5 km

3. His final week (Week 14) distance: 

a₁₄ = 5 + 13(1.5) = 5 + 19.5 = 24.5 km, which is well above 20 km.

Answer: True

4. aₙ > 20: 5 + (n−1)(1.5) > 20 

⇒ (n−1)(1.5) > 15 

⇒ n−1 > 10 

⇒ n > 11

Smallest whole n = 12. Check: a₁₂ = 5 + 11(1.5) = 5 + 16.5 = 21.5 km

Answer: Week 12

5. New S₁₄ = 14/2 [2(5) + 13(2)] = 7 [10 + 26] = 7 × 36 = 252 km

Difference = 252 − 206.5 = 45.5 km

Answer: 45.5 km more

Quick Revision

• aₙ = a + (n − 1)d

nth term of an AP

• Sₙ = n/2 [2a + (n − 1)d]

Sum of first n terms

• Sₙ = n/2 (a + l)

Alternate sum formula, l = last term

• d = aₙ − aₙ₋₁

Common difference

• aₘ = (aₘ₋ₖ + aₘ₊ₖ) / 2

Arithmetic mean property

Click below to download your free Case Study Questions PDF with worked-out examples for Class 10 Chapter 5:Arithmetic Progression, perfect for last-minute CBSE exam revision.

Class 10 Chapter 5: Arithmetic Progression Case Study PDF