Arithmetic Progressions PYQ Bank | Class X Maths

Questions

2-Mark Qs

3-Mark Qs

Years Covered

📋 Key Topics Covered (As per CBSE 2025-26 Syllabus)

nth Term Formula — aₙ = a + (n–1)d

Sum of n Terms — Sₙ = n/2 [2a + (n–1)d]

Sum (last term known) — Sₙ = n/2 (a + l)

Common Difference — d = a₂ – a₁

Finding k for AP — 2b = a + c condition

Word Problems — savings, instalments, patterns

✅

Full Chapter in Syllabus: All topics of Arithmetic Progressions are included in the 2025-26 syllabus. No deletions in this chapter. Algebra unit carries 20 marks — AP contributes 6-8 marks typically.

📝 SECTION A — 2-MARK QUESTIONS

15 Questions • Very Short Answer (VSA) Type • Board PYQs 2018–2025

CBSE 2025

The sequence –1, –1, –1, …, –1 is an AP. True or False? Justify. Also, if the common difference of an AP is 5, find a₁₈ – a₁₃.

2 Marks Assertion-Reason Type ⭐ MOST RECENT

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✅ DETAILED SOLUTION

Part 1 In the sequence –1, –1, –1, …, –1: d = –1 – (–1) = 0. Since the common difference is constant (d = 0), it is an AP. ✓ True

Part 2 a₁₈ – a₁₃ = [a + 17d] – [a + 12d] = 5d = 5 × 5 = 25

∴ True; a₁₈ – a₁₃ = 25

CBSE 2025

If the 2nd term of an AP is 13 and the 5th term is 25, then find its 7th term.

2 Marks nth Term ⭐ MOST RECENT

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✅ DETAILED SOLUTION

Step 1 a₂ = a + d = 13 … (i) ; a₅ = a + 4d = 25 … (ii)

Step 2 Subtracting (i) from (ii): 3d = 12 ⟹ d = 4

Step 3 From (i): a = 13 – 4 = 9

Step 4 a₇ = a + 6d = 9 + 24 = 33

∴ 7th term = 33

CBSE 2024

How many numbers lie between 10 and 300, which when divided by 4 leave a remainder 3? Also, find their sum.

2 Marks Count & Sum ⭐ IMPORTANT

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✅ DETAILED SOLUTION

Step 1 Numbers of the form 4k + 3: 11, 15, 19, …, 299 (AP with a = 11, d = 4, l = 299)

Step 2 n = (l – a)/d + 1 = (299 – 11)/4 + 1 = 288/4 + 1 = 73

Step 3 Sₙ = n/2 (a + l) = 73/2 (11 + 299) = 73/2 × 310 = 11,315

∴ 73 numbers; Sum = 11,315

CBSE 2023

If k+2, 4k–6, and 3k–2 are three consecutive terms of an AP, find the value of k.

2 Marks AP Condition ⭐ MOST REPEATED TYPE

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✅ DETAILED SOLUTION

For three terms in AP: 2(middle) = first + last

Step 1 2(4k – 6) = (k + 2) + (3k – 2)

Step 2 8k – 12 = 4k ⟹ 4k = 12 ⟹ k = 3

∴ k = 3

CBSE 2023

If a, b, c are in AP with common difference d, find the value of a – 2b – c.

2 Marks AP Property

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✅ DETAILED SOLUTION

Step 1 b = a + d, c = a + 2d

Step 2 a – 2b – c = a – 2(a + d) – (a + 2d) = a – 2a – 2d – a – 2d = –2a – 4d

∴ a – 2b – c = –2a – 4d

CBSE 2023

Which term of the AP: 65, 61, 57, 53, … is the first negative term?

2 Marks First Negative Term ⭐ CLASSIC

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✅ DETAILED SOLUTION

Step 1 a = 65, d = –4. We need aₙ < 0

Step 2 aₙ = 65 + (n–1)(–4) = 69 – 4n

Step 3 69 – 4n < 0 ⟹ n > 69/4 = 17.25 ⟹ n = 18

a₁₈ = 69 – 72 = –3 (first negative term)

∴ 18th term is the first negative term.

CBSE 2022

Find the value of x for which 2x, (x+10), and (3x+2) are the three consecutive terms of an AP.

2 Marks AP Condition

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✅ DETAILED SOLUTION

Step 1 For AP: 2(middle) = first + last ⟹ 2(x + 10) = 2x + (3x + 2)

Step 2 2x + 20 = 5x + 2 ⟹ 18 = 3x ⟹ x = 6

∴ x = 6 (Terms: 12, 16, 20 → d = 4 ✓)

CBSE 2022

How many terms are in the AP whose first and fifth terms are –14 and 2 respectively, and the last term is 62?

2 Marks Number of Terms

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✅ DETAILED SOLUTION

Step 1 a = –14, a₅ = a + 4d = 2 ⟹ –14 + 4d = 2 ⟹ d = 4

Step 2 Last term: aₙ = a + (n–1)d ⟹ 62 = –14 + (n–1)×4

Step 3 76 = 4(n–1) ⟹ n–1 = 19 ⟹ n = 20

∴ 20 terms

CBSE 2021

If the first term of an AP is –5 and the common difference is 2, then find the sum of the first 6 terms.

2 Marks Sum of n Terms

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✅ DETAILED SOLUTION

Step 1 a = –5, d = 2, n = 6

Step 2 S₆ = 6/2 [2(–5) + (6–1)(2)] = 3[–10 + 10] = 3 × 0 = 0

∴ S₆ = 0

Q10

CBSE 2021

What is the common difference of an AP in which a₂₁ – a₇ = 84?

2 Marks Common Difference

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✅ DETAILED SOLUTION

Step 1 a₂₁ – a₇ = (a + 20d) – (a + 6d) = 14d

Step 2 14d = 84 ⟹ d = 6

∴ Common difference d = 6

Q11

CBSE 2020

Which term of the AP: 21, 18, 15, … is –81? Also, is any term 0?

2 Marks nth Term

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✅ DETAILED SOLUTION

Step 1 a = 21, d = –3. Let aₙ = –81

Step 2 21 + (n–1)(–3) = –81 ⟹ –3(n–1) = –102 ⟹ n–1 = 34 ⟹ n = 35

Step 3 Is any term 0? 21 + (n–1)(–3) = 0 ⟹ n–1 = 7 ⟹ n = 8. Yes, 8th term is 0.

∴ –81 is the 35th term; Yes, the 8th term is 0.

Q12

CBSE 2019

Find the common difference of the AP: 1/a, (3–a)/3a, (3–2a)/3a, … (a ≠ 0)

2 Marks Common Difference

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✅ DETAILED SOLUTION

Step 1 First term = 1/a = 3/3a

Step 2 d = (3–a)/3a – 3/3a = (3–a–3)/3a = –a/3a = –1/3

∴ Common difference d = –1/3

Q13

CBSE 2020

If Sₙ = 4n – n² for an AP, find the AP (i.e., first few terms).

2 Marks Sₙ → AP

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✅ DETAILED SOLUTION

Step 1 a₁ = S₁ = 4(1) – 1² = 3

Step 2 a₂ = S₂ – S₁ = (8 – 4) – 3 = 1

Step 3 d = a₂ – a₁ = 1 – 3 = –2

∴ AP: 3, 1, –1, –3, –5, … (a = 3, d = –2)

Q14

CBSE 2018

How many multiples of 4 lie between 10 and 250?

2 Marks Count Terms

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✅ DETAILED SOLUTION

Step 1 Multiples of 4 between 10 and 250: 12, 16, 20, …, 248

Step 2 a = 12, d = 4, l = 248

Step 3 n = (248 – 12)/4 + 1 = 236/4 + 1 = 59 + 1 = 60

∴ 60 multiples of 4 lie between 10 and 250.

Q15

CBSE 2019

Ramkali saved ₹5 in the first week and increased her saving by ₹1.75 each week. If in the nth week her saving becomes ₹20.75, find n.

2 Marks Word Problem

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✅ DETAILED SOLUTION

Step 1 a = 5, d = 1.75, aₙ = 20.75

Step 2 20.75 = 5 + (n–1) × 1.75 ⟹ 15.75 = 1.75(n–1)

Step 3 n–1 = 15.75/1.75 = 9 ⟹ n = 10

∴ n = 10 (10th week)

📝 SECTION B — 3-MARK QUESTIONS

15 Questions • Short Answer (SA) Type • Board PYQs 2018–2025

Q16

CBSE 2024

The ratio of the 10th term to the 30th term of an AP is 1 : 3 and the sum of its first six terms is 42. Find the first term and the common difference.

3 Marks Find a and d ⭐ MOST IMPORTANT

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✅ DETAILED SOLUTION

Step 1 a₁₀/a₃₀ = 1/3 ⟹ (a+9d)/(a+29d) = 1/3

3(a+9d) = a+29d ⟹ 3a+27d = a+29d ⟹ 2a = 2d ⟹ a = d … (i)

Step 2 S₆ = 42 ⟹ 6/2 [2a + 5d] = 42 ⟹ 3(2a+5d) = 42 ⟹ 2a+5d = 14

Step 3 From (i): 2a + 5a = 14 ⟹ 7a = 14 ⟹ a = 2, d = 2

∴ First term a = 2, Common difference d = 2

Q17

CBSE 2024

If the sum of the first 7 terms of an AP is 49 and that of the first 17 terms is 289, find the sum of its first 20 terms.

3 Marks Sum of n Terms ⭐ HOTS

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✅ DETAILED SOLUTION

Step 1 S₇ = 49 ⟹ 7/2 [2a + 6d] = 49 ⟹ 2a + 6d = 14 ⟹ a + 3d = 7 … (i)

Step 2 S₁₇ = 289 ⟹ 17/2 [2a + 16d] = 289 ⟹ 2a + 16d = 34 ⟹ a + 8d = 17 … (ii)

Step 3 (ii) – (i): 5d = 10 ⟹ d = 2, a = 7 – 6 = 1

Step 4 S₂₀ = 20/2 [2(1) + 19(2)] = 10 × 40 = 400

∴ Sum of first 20 terms = 400

Q18

CBSE 2024

If pth term of an AP is q and qth term is p, then prove that its nth term is (p + q – n).

3 Marks Prove ⭐ IMPORTANT

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✅ DETAILED SOLUTION

Step 1 aₚ = a + (p–1)d = q … (i) ; a_q = a + (q–1)d = p … (ii)

Step 2 (i) – (ii): (p–q)d = q – p ⟹ d = –1

Step 3 From (i): a = q – (p–1)(–1) = q + p – 1

Step 4 aₙ = a + (n–1)d = (p+q–1) + (n–1)(–1) = p + q – 1 – n + 1 = p + q – n

∴ nth term = p + q – n. Hence proved.

Q19

CBSE 2023

Solve: (–4) + (–1) + 2 + 5 + … + x = 437

3 Marks Find Last Term ⭐ HOTS

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✅ DETAILED SOLUTION

Step 1 a = –4, d = 3. Let the number of terms = n, last term = x

Step 2 Sₙ = n/2 [2a + (n–1)d] = n/2 [–8 + 3n – 3] = n(3n–11)/2 = 437

Step 3 3n² – 11n – 874 = 0. Using quadratic formula:

n = [11 ± √(121 + 10488)] / 6 = [11 ± √10609] / 6 = [11 ± 103] / 6

Step 4 n = 114/6 = 19 (reject negative value)

Step 5 x = a + (n–1)d = –4 + 18×3 = –4 + 54 = 50

∴ x = 50

Q20

CBSE 2023

The ratio of the 11th term to the 18th term of an AP is 2 : 3. Find the ratio of the 5th term to the 21st term, and the ratio of the sum of first 5 terms to the sum of first 21 terms.

3 Marks Ratio of Terms & Sums ⭐ CHALLENGING

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✅ DETAILED SOLUTION

Step 1 a₁₁/a₁₈ = 2/3 ⟹ (a+10d)/(a+17d) = 2/3

3a + 30d = 2a + 34d ⟹ a = 4d

Step 2 a₅/a₂₁ = (a+4d)/(a+20d) = (4d+4d)/(4d+20d) = 8d/24d = 1/3

Step 3 S₅ = 5/2[2a + 4d] = 5/2[8d + 4d] = 5/2 × 12d = 30d

S₂₁ = 21/2[2a + 20d] = 21/2[8d + 20d] = 21/2 × 28d = 294d

Step 4 S₅/S₂₁ = 30d/294d = 5/49

∴ a₅ : a₂₁ = 1 : 3 ; S₅ : S₂₁ = 5 : 49

Q21

CBSE 2023

Rohan takes a loan of ₹1,18,000 and pays every month starting with ₹1,000 and increases the instalment by ₹100 each month. What amount will be paid in the 30th instalment? What total loan has he paid after 30 instalments?

3 Marks Instalment Problem ⭐ APPLICATION BASED

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✅ DETAILED SOLUTION

Step 1 AP: 1000, 1100, 1200, … (a = 1000, d = 100)

Step 2 a₃₀ = 1000 + 29 × 100 = 1000 + 2900 = ₹3,900

Step 3 S₃₀ = 30/2 (1000 + 3900) = 15 × 4900 = ₹73,500

∴ 30th instalment = ₹3,900 ; Total paid after 30 instalments = ₹73,500

Q22

CBSE 2022

In an AP, the 24th term is twice the 10th term. Prove that the 36th term is twice the 16th term.

3 Marks Prove

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✅ DETAILED SOLUTION

Step 1 Given: a₂₄ = 2a₁₀ ⟹ a + 23d = 2(a + 9d) = 2a + 18d

⟹ 23d – 18d = a ⟹ a = 5d … (i)

Step 2 a₃₆ = a + 35d = 5d + 35d = 40d

Step 3 a₁₆ = a + 15d = 5d + 15d = 20d

Step 4 a₃₆ = 40d = 2 × 20d = 2 × a₁₆

∴ a₃₆ = 2 × a₁₆. Hence proved.

Q23

CBSE 2021

How many terms of the AP: 24, 21, 18, … must be taken so that their sum is 78? Explain the double answer.

3 Marks Sum → Find n ⭐ CLASSIC

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✅ DETAILED SOLUTION

Step 1 a = 24, d = –3. Sₙ = 78

Step 2 n/2 [2(24) + (n–1)(–3)] = 78 ⟹ n/2 [48 – 3n + 3] = 78

n(51 – 3n) = 156 ⟹ 3n² – 51n + 156 = 0 ⟹ n² – 17n + 52 = 0

Step 3 (n – 4)(n – 13) = 0 ⟹ n = 4 or n = 13

Why two answers? After the 8th term (a₈ = 0), terms become negative. Sum of terms from 5th to 13th is zero! So S₄ = S₁₃ = 78. Both answers are valid.

∴ n = 4 or n = 13

Q24

CBSE 2020

The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.

3 Marks Find n and d

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✅ DETAILED SOLUTION

Step 1 a = 5, l = 45, S = 400

Step 2 S = n/2 (a + l) ⟹ 400 = n/2 (5 + 45) = 25n ⟹ n = 16

Step 3 l = a + (n–1)d ⟹ 45 = 5 + 15d ⟹ 15d = 40 ⟹ d = 8/3

∴ n = 16, d = 8/3

Q25

CBSE 2022

The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.

3 Marks Find AP ⭐ IMPORTANT

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✅ DETAILED SOLUTION

Step 1 a₄ + a₈ = 24 ⟹ (a+3d) + (a+7d) = 24 ⟹ 2a+10d = 24 ⟹ a+5d = 12 … (i)

Step 2 a₆ + a₁₀ = 44 ⟹ (a+5d) + (a+9d) = 44 ⟹ 2a+14d = 44 ⟹ a+7d = 22 … (ii)

Step 3 (ii) – (i): 2d = 10 ⟹ d = 5. From (i): a = 12–25 = –13

Step 4 a₂ = –13+5 = –8, a₃ = –8+5 = –3

∴ First three terms: –13, –8, –3

Q26

CBSE 2019

Ramkali required ₹5,000 to get her daughter admitted to school after a year. She saved ₹150 in the first month and increased by ₹50 every month. Can she arrange ₹5,000 after 12 months?

3 Marks Savings Problem ⭐ VALUE BASED

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✅ DETAILED SOLUTION

Step 1 Savings: 150, 200, 250, 300, … (a = 150, d = 50, n = 12)

Step 2 S₁₂ = 12/2 [2(150) + 11(50)] = 6[300 + 550] = 6 × 850 = ₹5,100

Step 3 ₹5,100 > ₹5,000 ✓

Value: Ramkali's efforts reflect her awareness about girls' education and the importance of educating her child.

∴ Yes, she can arrange ₹5,000 (total savings = ₹5,100).

Q27

CBSE 2021

Find the sum of the first 22 terms of an AP in which d = 7 and the 22nd term is 149.

3 Marks Sum of n Terms

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✅ DETAILED SOLUTION

Step 1 a₂₂ = 149, d = 7 ⟹ a + 21(7) = 149 ⟹ a = 149 – 147 = 2

Step 2 S₂₂ = 22/2 (a + a₂₂) = 11(2 + 149) = 11 × 151 = 1,661

∴ Sum of first 22 terms = 1,661

Q28

CBSE 2020

Find the sum of all multiples of 9 lying between 100 and 200.

3 Marks Sum of Multiples

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✅ DETAILED SOLUTION

Step 1 Multiples of 9 between 100 and 200: 108, 117, 126, …, 198

Step 2 a = 108, d = 9, l = 198. n = (198–108)/9 + 1 = 90/9 + 1 = 11

Step 3 S₁₁ = 11/2 (108 + 198) = 11/2 × 306 = 1,683

∴ Sum = 1,683

Q29

CBSE 2018

A sum of ₹700 is to be paid as seven cash prizes to students. If each prize is ₹20 less than the preceding one, find the value of each prize.

3 Marks Prize Distribution

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✅ DETAILED SOLUTION

Step 1 Let first (highest) prize = a. AP: a, a–20, a–40, … (d = –20, n = 7)

Step 2 S₇ = 700 ⟹ 7/2 [2a + 6(–20)] = 700

7/2 [2a – 120] = 700 ⟹ 2a – 120 = 200 ⟹ a = 160

Step 3 Prizes: 160, 140, 120, 100, 80, 60, 40

Verification: 160+140+120+100+80+60+40 = 700 ✓

∴ Prizes: ₹160, ₹140, ₹120, ₹100, ₹80, ₹60, ₹40

Q30

CBSE 2018

The sum of the first 12 terms of an AP is 900. If its first term is 20, find the common difference and the 12th term.

3 Marks Find d and a₁₂

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✅ DETAILED SOLUTION

Step 1 a = 20, n = 12, S₁₂ = 900

Step 2 12/2 [2(20) + 11d] = 900 ⟹ 6[40 + 11d] = 900 ⟹ 40 + 11d = 150

Step 3 11d = 110 ⟹ d = 10

Step 4 a₁₂ = 20 + 11(10) = 20 + 110 = 130

∴ d = 10, 12th term = 130

📐 KEY FORMULAS TO REMEMBER

General (nth) Term: aₙ = a + (n – 1)d

Sum of n Terms: Sₙ = n/2 [2a + (n – 1)d]

Sum (last term known): Sₙ = n/2 (a + l)

nth term from Sₙ: aₙ = Sₙ – Sₙ₋₁ (for n ≥ 2), a₁ = S₁

AP Condition (3 terms): 2b = a + c
Common Difference: d = a₂ – a₁ = aₙ – aₙ₋₁
Number of Terms: n = (l – a)/d + 1
Middle Term: For odd n terms, middle term = a + [(n–1)/2]d
Sum of first n natural numbers: n(n+1)/2
aₘ – aₙ = (m – n)d (useful shortcut!)

💡 BOARD EXAM TIPS — ARITHMETIC PROGRESSIONS

"Find k for AP" trick: Use 2(middle) = first + last. Saves time on MCQs & 2-mark Qs.

aₘ – aₙ shortcut: Directly = (m – n) × d. No need to find a first!

Double answer in sum problems: If the AP has negative terms, Sₙ can repeat. Always explain WHY both values are valid.

Word problems (savings, instalments): Identify a, d, n clearly. Use Sₙ = n/2(a+l) when last term is known — it's faster!

Expected marks: 6–8 marks from this chapter (MCQ + SA + possible Case Study).

Always verify: Quick-check your answer by computing the first few terms and cross-checking the sum.