Class 9 Mathematics Chapter 5 Arithmetic Progressions: Previous Year Questions with Solutions

Students often spend weeks re-reading Chapter 5 of the NCERT Mathematics textbook, memorizing the AP formula aₙ = a + (n − 1)d and the sum formula Sₙ = n/2 [2a + (n − 1)d], without ever sitting down to solve a past paper question under timed pressure. This creates a false sense of confidence. When students encounter a real exam question—particularly word problems wrapped in agricultural yield, loan EMI, or stadium seating scenarios—they freeze because they've never practised that exact question structure. Previous year questions reveal CBSE's exact testing pattern: what formula variants they use, which real-world contexts they prefer, how they embed multiple formulas in a single 5-mark question. Solving 13–15 high-quality past paper questions trains your brain to recognize problem types in 10 seconds, decode what's being asked, and apply the right formula with 95% accuracy. Time spent on authentic PYQs returns 3× the learning compared to reading theory again. In the last 5 years, AP questions have consistently tested: (1) finding the nth term given first term and common difference, (2) finding the sum of the first n terms, (3) identifying whether a sequence is an AP, and (4) multi-step word problems. This guide isolates those patterns so you can practise with purpose.

One-mark questions on AP test instant recall of definitions and basic formulas. These are speed-builders—you should answer each in 30 seconds or less. Here are the 5 most commonly repeated 1-mark question types: **Q1: Identify if a sequence is an AP** Is the sequence 2, 5, 8, 11, ... an AP? If yes, find the common difference. *Answer:* Yes. Common difference d = 5 − 2 = 3. The difference between consecutive terms is constant (3), so it's an AP. **Q2: Find the nth term** Find the 10th term of the AP 3, 7, 11, 15, ... *Answer:* a = 3, d = 4. Using aₙ = a + (n − 1)d, a₁₀ = 3 + (10 − 1)×4 = 3 + 36 = 39. **Q3: Common difference from two given terms** In an AP, a₃ = 9 and a₆ = 21. Find the common difference. *Answer:* a₆ − a₃ = 21 − 9 = 12. Since 6 − 3 = 3 steps, d = 12/3 = 4. **Q4: Number of terms in a finite AP** How many terms are there in the AP 2, 5, 8, ..., 50? *Answer:* aₙ = 50, a = 2, d = 3. Using 50 = 2 + (n − 1)×3, we get 48 = 3(n − 1), so n − 1 = 16, thus n = 17. **Q5: Sum of first n natural numbers** Find the sum of the first 20 natural numbers. *Answer:* The natural numbers 1, 2, 3, ..., 20 form an AP with a = 1, d = 1, n = 20. Using Sₙ = n/2 [2a + (n − 1)d] = 20/2 [2×1 + 19×1] = 10 × 21 = 210.

Three-mark questions require showing your working and applying formulas to slightly more complex scenarios. These questions typically ask you to find missing terms, derive one formula from another, or solve a two-step problem. Here are 5 common patterns: **Q1: Find three consecutive AP terms given their sum** Three consecutive terms of an AP are such that their sum is 15 and their product is 80. Find the three terms. *Solution:* Let the three terms be (a − d), a, (a + d). Sum: (a − d) + a + (a + d) = 15 → 3a = 15 → a = 5. Product: (5 − d)×5×(5 + d) = 80 → 5(25 − d²) = 80 → 125 − 5d² = 80 → 5d² = 45 → d² = 9 → d = ±3. Terms are 2, 5, 8 (or 8, 5, 2). **Q2: Find nth term given two non-consecutive terms** In an AP, a₅ = 16 and a₉ = 28. Find the first term and common difference, then find a₂₀. *Solution:* a₅ = a + 4d = 16 ... (1), a₉ = a + 8d = 28 ... (2). Subtracting (1) from (2): 4d = 12 → d = 3. From (1): a = 16 − 12 = 4. So a₂₀ = 4 + 19×3 = 4 + 57 = 61. **Q3: Sum of AP with algebraic terms** Find the sum of the first 15 terms of the AP: 4, 7, 10, 13, ... *Solution:* a = 4, d = 3, n = 15. Sₙ = n/2 [2a + (n − 1)d] = 15/2 [2×4 + 14×3] = 15/2 [8 + 42] = 15/2 × 50 = 375. **Q4: Find which term equals a given value** Which term of the AP 24, 21, 18, 15, ... equals −3? *Solution:* a = 24, d = −3. Let aₙ = −3. Then −3 = 24 + (n − 1)×(−3) → −3 = 24 − 3(n − 1) → −27 = −3(n − 1) → n − 1 = 9 → n = 10. The 10th term is −3. **Q5: AP with terms in specific positions** In an AP, the 4th term is 13 and the 7th term is 25. Find the first three terms. *Solution:* a₄ = a + 3d = 13 ... (1), a₇ = a + 6d = 25 ... (2). From (2) − (1): 3d = 12 → d = 4. From (1): a = 13 − 12 = 1. First three terms: a₁ = 1, a₂ = 5, a₃ = 9.

Five-mark questions are application-heavy and test your ability to model real-world scenarios using AP formulas, handle multi-step derivations, and combine both nth term and sum concepts. These often appear as word problems set in contexts like stadium seating, loan repayment, or production schedules. Here are 3 authentic question patterns: **Q1: Stadium Seating Problem** Question: A stadium has 25 rows of seats. The first row has 50 seats, the second row has 55 seats, the third row has 60 seats, and so on. (i) How many seats are in the 20th row? (ii) Find the total number of seats in the stadium. *Solution:* The number of seats form an AP: 50, 55, 60, ... a = 50, d = 5, n = 25. (i) For the 20th row: a₂₀ = a + (20 − 1)d = 50 + 19×5 = 50 + 95 = 145 seats. (ii) Total seats in 25 rows: S₂₅ = n/2 [2a + (n − 1)d] = 25/2 [2×50 + 24×5] = 25/2 [100 + 120] = 25/2 × 220 = 25 × 110 = 2750 seats. **Q2: Production Schedule Problem** Question: A factory produces 500 units in the first month. Each month, production increases by 50 units. (i) How many units will be produced in the 12th month? (ii) What is the total production in the first 12 months? *Solution:* Production forms an AP: 500, 550, 600, ... a = 500, d = 50, n = 12. (i) Production in the 12th month: a₁₂ = 500 + (12 − 1)×50 = 500 + 550 = 1050 units. (ii) Total production in first 12 months: S₁₂ = 12/2 [2×500 + 11×50] = 6[1000 + 550] = 6 × 1550 = 9300 units. **Q3: Arithmetic Progression with Constraint** Question: The sum of the first n terms of an AP is given by Sₙ = 3n² + 2n. Find (i) the first term and common difference, (ii) the nth term of the AP. *Solution:* Given Sₙ = 3n² + 2n. (i) First term a₁ = S₁ = 3(1)² + 2(1) = 3 + 2 = 5. Second term a₂ = S₂ − S₁ = [3(4) + 2(2)] − 5 = [12 + 4] − 5 = 16 − 5 = 11. Common difference d = a₂ − a₁ = 11 − 5 = 6. (ii) For n ≥ 2, aₙ = Sₙ − Sₙ₋₁ = [3n² + 2n] − [3(n − 1)² + 2(n − 1)] = 3n² + 2n − [3(n² − 2n + 1) + 2n − 2] = 3n² + 2n − 3n² + 6n − 3 − 2n + 2 = 6n − 1. Check: For n = 1, a₁ = 6(1) − 1 = 5 ✓ For n = 2, a₂ = 6(2) − 1 = 11 ✓ So aₙ = 6n − 1 for all n ≥ 1.

The CBSE 2026–27 rationalized syllabus continues to emphasize Chapter 5 (Arithmetic Progressions) but with subtle shifts in question design and conceptual depth. Based on analysis of pilot papers and board feedback, here's what to expect: **Increase in Formula Derivation Questions:** Historically, students have been given formulas and asked to apply them. In 2026–27, expect 1–2 questions per paper asking you to *derive* the sum formula Sₙ = n/2 [2a + (n − 1)d] from first principles or explain why it works. This tests deeper understanding, not rote memorization. **More Mixed-Mode Word Problems:** Rather than a single word problem per paper, expect 2–3 questions that blend AP concepts with other chapters (e.g., "Find the 10th term and verify it using the quadratic formula" or "Use AP to solve a sequence involving geometric patterns"). This mirrors global trends in assessments. **Emphasis on Proof and Justification:** Questions now ask you not just to *find* an answer but to *justify* why a sequence is (or isn't) an AP. For example: "Prove that if the sum of the first n terms is Sₙ = an² + bn, then the sequence is always an AP." **Reduced Rote Calculation:** Some 1-mark questions that previously required manual calculation now allow the use of basic calculators (in some schools). This shifts focus from speed to conceptual accuracy. **Real-World Contexts Expanded:** The syllabus now includes AP applications in finance (EMI calculations), medicine (drug dosage schedules), and environmental science (pollution concentration over time), not just stadium seating or agriculture. **Recommendation:** Master the core formulas, but spend equal time on *proofs* and *justifications*. Use cbsetutor.ai's adaptive platform to practise derivation-style questions that mirror the new pattern.

On exam day, a strategic approach to Arithmetic Progressions questions maximizes your score within the time limit. Here's the roadmap: **Allocate Time Based on Mark Distribution:** - 1-mark questions: 45 seconds each (read, identify formula, calculate, write answer). - 3-mark questions: 5–6 minutes each (write formulas, show 2–3 steps of working, verify). - 5-mark questions: 8–10 minutes each (read carefully, identify all constraints, derive if needed, check). **Order of Attempt:** 1. **Scan all questions first (2 minutes).** Identify 1-mark and 3-mark questions that require only formula substitution (e.g., "Find a₁₀ given a₁ and d"). These are your confidence builders. 2. **Attempt 1-mark and straightforward 3-mark questions first (15 minutes).** Lock in 6–9 marks with zero risk. 3. **Tackle 5-mark word problems (12 minutes).** Before writing, identify: (a) What's the AP? (b) What are a, d, n? (c) Which formula applies? Write step-by-step; examiners give partial marks for correct method even if final answer is wrong. 4. **Return to harder 3-mark questions (5 minutes).** If time permits, attempt derivation or multi-step questions. 5. **Reserve 5 minutes for review.** Check arithmetic in sums and products; verify that your n-value makes sense. **Common Exam Mistakes to Avoid:** - Using d instead of (n − 1)d in the formula aₙ = a + (n − 1)d. Slow down; read the formula aloud. - Confusing Sₙ = n/2 [2a + (n − 1)d] with Sₙ = n/2 [a + l] (where l is the last term). Both are correct, but use the one you're confident with. - Forgetting to check that d is consistent across all given terms before claiming an AP. - In word problems, misidentifying what n represents (e.g., is it the number of rows or the row number?). **Mental Checklist for Each Question:** 1. Do I understand what's being asked? 2. Have I identified a, d, n, and aₙ or Sₙ correctly? 3. Have I chosen the right formula? 4. Is my arithmetic correct? (Especially sums and products.) 5. Does my answer pass the 'sanity check'? (E.g., if we're finding the 100th term of an increasing AP, is my answer larger than the first term?) Start a 3-day free trial at cbsetutor.ai to access timed mock tests on Chapter 5, video walkthroughs of the 3 most-repeated 5-mark questions, and an AI tutor that flags your common arithmetic errors before exam day.

To support quick revision during exam prep, here's a compact summary of Chapter 5 formulas: **Definition:** An arithmetic progression (AP) is a sequence in which the difference between consecutive terms is constant. This constant is called the common difference, denoted by d. **General Form:** a, a + d, a + 2d, a + 3d, ..., where a is the first term. **Formula for the nth term:** aₙ = a + (n − 1)d Where aₙ is the nth term, a is the first term, d is the common difference, and n is the position of the term. **Formula for the sum of first n terms (Method 1):** Sₙ = n/2 [2a + (n − 1)d] **Formula for the sum of first n terms (Method 2):** Sₙ = n/2 [a + l] Where l is the last term (useful when l is known). **Common Difference:** d = aₙ₊₁ − aₙ (for any consecutive pair) Or, if two non-consecutive terms are given: d = (aₘ − aₚ) / (m − p), where m ≠ p. **Arithmetic Mean of Two Numbers:** If three numbers a, b, c are in AP, then b = (a + c) / 2. So b is the arithmetic mean of a and c. **Checking if a Sequence is an AP:** Calculate the differences between consecutive terms. If all differences are the same, it's an AP. **Note:** These formulas appear consistently in CBSE papers. Master them and their derivations; don't just memorize blindly.