CBSE
Class 10 Arithmetic Progression Case Study Questions, Solved
CBSE case study questions give a real-life setting, then three or four sub-parts that all use an = a + (n - 1)d or Sn = n/2 × (2a + (n - 1)d). Below are three original scenarios, theatre seats, a monthly savings plan and ladder rungs, each solved sub-part by sub-part with the formula stated first.
Since CBSE moved towards competency-based papers, the arithmetic progression chapter almost always shows up as a case study: a paragraph about seats, savings or steps, followed by three or four short sub-parts. Past paper questions cannot be reproduced here, so the three scenarios below, a theatre, a monthly savings plan and a carpenter's ladder, are original, but they are built to the same shape and difficulty as the real thing. Each sub-part is solved with the formula stated before the substitution, the way marking schemes reward. If you can do these three cold, the exam version should not surprise you.
What the case study format looks like
A case study question opens with a short real-world paragraph, seats in a stadium, instalments on a loan, steps of a staircase, and then asks three or four sub-parts that escalate: typically identify a or d, find a particular term, then find a sum, sometimes with a small twist in the last part. The sub-parts are marked separately, usually adding to 4 or 5 marks, so a blank part (iii) does not erase your (i) and (ii). The scenarios on this page are original, because real past-paper questions cannot be republished, but they follow the same escalation and the same mark shape. Train yourself to extract three things from the paragraph before answering anything: the first term, the common difference, and what n means in this particular story.
Three formulas, and how to pick between the two sums
Everything here runs on the nth term formula an = a + (n - 1)d and the two sum formulas Sn = n/2 × (2a + (n - 1)d) and Sn = n/2 × (a + l). The nth term answers questions about one position: how many seats in row 10, one deposit, one rung. The sums answer questions about totals: capacity, savings so far, wood needed. Choosing between the two sum forms is simple: if you already know the last term l, as in the ladder scenario where both end rungs are given, use n/2 × (a + l) and finish in one line; if you only know a and d, use the longer form. Writing the formula before substituting is worth an explicit mark in most schemes, so never skip that line even under time pressure.
Error carried forward, and why sub-parts reward method
Case studies chain sub-parts, so students worry that a slip in part (i) poisons everything after it. In practice CBSE marking generally applies error carried forward: if your d is wrong but your part (ii) method correctly uses your d, the method credit in (ii) survives. That changes how you should play the question. Never abandon later sub-parts because an earlier answer feels doubtful, and never erase working, because visible method is what the carried-forward credit attaches to. Do sanity-check magnitudes as you go: 30 rows starting at 12 seats cannot plausibly hold 40,000 people, and a negative number of seats means a sign slipped somewhere. A ten-second reality check between sub-parts catches most disasters while they are still worth only one mark.
Worked questions, step by step
Case study: A new theatre hall has 30 rows of seats. The first row, nearest the stage, has 12 seats, and every row after it has 3 seats more than the row in front of it. (i) How many seats are in the 10th row? (ii) Which row has exactly 57 seats? (iii) How many seats does the theatre hold in total?
- Given: the seat counts form an AP with first term a = 12, common difference d = 3, and n = 30 rows in all.
- (i) Formula: an = a + (n - 1)d, so a10 = 12 + 9 × 3 = 12 + 27 = 39 seats.
- (ii) Set an = 57: 12 + (n - 1) × 3 = 57, so (n - 1) × 3 = 45, n - 1 = 15 and n = 16.
- (iii) Formula: Sn = n/2 × (2a + (n - 1)d), so S30 = 30/2 × (2 × 12 + 29 × 3).
- S30 = 15 × (24 + 87) = 15 × 111 = 1665.
- Answer: (i) 39 seats, (ii) the 16th row, (iii) 1,665 seats in total.
Answer: (i) 39 seats (ii) 16th row (iii) 1,665 seats
Where marks slip: In part (ii) students solve (n - 1) × 3 = 45 correctly and then forget the final plus one, answering row 15; the n you want is 15 + 1 = 16, so always finish the rearrangement.
Try one yourself: An open-air theatre has 25 rows, with 10 seats in the first row and 2 extra seats in each row after it. Find the seats in the last row and the total capacity. (Answer: 58 seats and 850 seats)
Case study: Priya opens a savings account in January and deposits Rs 200 that month. Each month after that, she deposits Rs 50 more than she did the month before. (i) How much does she deposit in the 12th month? (ii) In which month does she deposit exactly Rs 600? (iii) How much has she saved in total after one full year?
- Given: the deposits form an AP with a = 200 and d = 50. To find: the 12th deposit, the month of the Rs 600 deposit, and the 12-month total.
- (i) Formula: an = a + (n - 1)d, so a12 = 200 + 11 × 50 = 200 + 550 = Rs 750.
- (ii) Set an = 600: 200 + (n - 1) × 50 = 600, so (n - 1) × 50 = 400 and n = 9, the 9th month.
- (iii) Formula: Sn = n/2 × (2a + (n - 1)d), so S12 = 12/2 × (2 × 200 + 11 × 50) = 6 × (400 + 550).
- S12 = 6 × 950 = Rs 5,700.
- Answer: (i) Rs 750, (ii) the 9th month, (iii) Rs 5,700 saved over the year.
Answer: (i) Rs 750 (ii) 9th month (iii) Rs 5,700
Where marks slip: Mixing up an and Sn is the killer here: part (i) asks for one deposit, the nth term, while part (iii) asks for the running total, the sum, so label which formula you are using before substituting.
Try one yourself: Arjun saves Rs 150 in the first month and increases his saving by Rs 25 each month. How much has he saved in total after 24 months? (Answer: Rs 10,500)
Case study: A carpenter is building a ladder with 15 rungs whose lengths decrease evenly from bottom to top. The bottom rung is 60 cm long and the top rung is 32 cm long. (i) Explain why the rung lengths form an AP and find the common difference. (ii) How long is the 8th rung from the bottom? (iii) What total length of wood is needed for all 15 rungs?
- Given: 15 rungs, first term a = 60 cm, last term l = 32 cm, with the lengths decreasing evenly.
- (i) Because the lengths change by the same amount each step, they form an AP; d = (l - a)/(n - 1) = (32 - 60)/14 = -28/14 = -2 cm.
- (ii) Formula: an = a + (n - 1)d, so a8 = 60 + 7 × (-2) = 60 - 14 = 46 cm.
- (iii) Since the last term is known, use Sn = n/2 × (a + l): S15 = 15/2 × (60 + 32).
- S15 = 15/2 × 92 = 15 × 46 = 690 cm.
- Answer: (i) d = -2 cm, (ii) 46 cm, (iii) 690 cm of wood, which is 6.9 m.
Answer: (i) d = -2 cm (ii) 46 cm (iii) 690 cm (6.9 m)
Where marks slip: When both the first and last terms are given, Sn = n/2 × (a + l) finishes in one line; students who insist on the 2a + (n - 1)d version often slip on the negative d, so pick the shorter formula.
Try one yourself: A ladder has 11 rungs shrinking evenly from 50 cm at the bottom to 30 cm at the top. Find the total length of wood in the rungs. (Answer: 440 cm)
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Solve my question →Quiz me on this topicQuestions students ask
Are case study questions compulsory in the Class 10 maths paper?
Recent CBSE papers include a dedicated case-based section, and arithmetic progression is one of its most frequent settings, so treat it as effectively certain rather than optional. The exact structure, typically 4 marks split across three or four sub-parts, sometimes with internal choice, is set each year in the official sample paper, which is worth checking before the exam.
Which formulas do AP case studies actually use?
Three: the nth term an = a + (n - 1)d for any single-position question, and the sums Sn = n/2 × (2a + (n - 1)d) or Sn = n/2 × (a + l) for totals. Case studies rarely go beyond these three; the difficulty is extracting a, d and n from the story, not the algebra itself.
If I get part (i) wrong, are the later parts lost too?
Generally no. CBSE marking normally applies error carried forward, so a later sub-part solved with a correct method using your earlier wrong value still collects its method credit. This only works if your working is visible, so write every formula and substitution out and never erase an attempt, even when you suspect the number feeding into it.
Are these real past paper questions?
No. Past CBSE questions are copyrighted and cannot be republished, so all three scenarios here are original, written to match the pattern, difficulty and mark structure of the real case studies. For exam-authentic wording, pair this page with the official CBSE sample papers for the current year, which are free on the board's website.