Arithmetic Progression: Important Questions with Step-by-Step Solutions for Class 10

Arithmetic Progression (AP) is an important chapter in Class 10 Mathematics that introduces students to number patterns and sequences with a constant difference between consecutive terms. This topic helps students understand concepts such as finding the nth term of an AP, determining the common difference, and calculating the sum of finite terms using standard formulas. In this guide, you will find carefully selected important questions covering all major concepts from the Class 10 Arithmetic Progression chapter, designed according to the latest syllabus and exam pattern to support effective preparation and revision.

Our subject experts have provided detailed solutions for these problems based on the latest CBSE syllabus and the NCERT textbook. This material helps students revise the chapter easily and perform well in the final examination.

Table of Contents

• All Important Arithmetic Progression Formulae
• Section A: 1-Mark Questions (MCQ & VSA)
• Section B: 2-Mark Questions (Short Answer)
• Section C: 3-Mark Questions
• Section D: Word Problems & Real Life Applications of Arithmetic Progression
• Section E: 4 & 5-Mark Questions with Case Studies
• Previous Year Questions on Arithmetic Progression

All Important Arithmetic Progression Formulae

 
An arithmetic progression (AP) is a sequence a, a+d, a+2d, a+3d, … where each term is obtained by adding a fixed number d to the previous term.

First Term (a):

The very first number in the sequence. Sometimes written as a₁. 

a = a₁ (1st term)

Common Difference (d):

Common difference is the constant gap between consecutive terms. d can be positive (increasing AP), zero (constant AP), or negative (decreasing AP).

d = a₂ − a₁ = a₃ − a₂

Section A: 1-Mark Questions (MCQ & VSA)

Q1: Three numbers in AP have the sum 30. What is the middle term?

(a) 4 (b) 10 

(c) 16 (d) 8

Solution: Let the three terms be a−d, a, a+d. 

Their sum = 3a = 30 

⇒ a = 10. 

The middle term is 10.

Answer: (b) 10 

Q2: Which term of the AP 3, 8, 13, 18, … is 78?

(a) 14 (b) 13

(c) 16 (d) 15

Solution: a = 3, d = 5. 

aₙ = 78 

⇒ 3 + (n−1)×5 = 78 

⇒ 5(n−1) = 75 

⇒ n−1 = 15 

⇒ n = 16.

Answer: (c) 16 

Q3: If the common difference d = 0 in an AP, what is the nature of the sequence?

(a) Increasing sequence (b) Decreasing sequence

(c) Constant sequence  (d) Not an AP

Solution: d = 0 means every term equals the first term. Example: 5, 5, 5, 5, … All terms are constant.

Q4: Find the 11th term of the AP: −5, −5/2, 0, 5/2, …

Solution: a = −5, d = −5/2 − (−5) = 5/2. 

aₙ = a + (n−1)d

a₁₁ = −5 + 10 × (5/2) = −5 + 25 = 20

a₁₁ = 20

Q5: Find the common difference of the AP: 1/a, (3−a)/3a, (3−2a)/3a, … (a ≠ 0)

Solution: a₂ − a₁ = (3−a)/3a − 1/a 

= (3−a)/3a − 3/3a 

= (3−a−3)/3a 

= −a/3a = −1/3

Common difference d = −1/3

Q6: Is −100 a term of the AP 11, 8, 5, 2, …?

Solution: a = 11, d = −3. 

If −100 is a term: 11 + (n−1)(−3) = −100 

⇒ (n−1) = −111/(−3) = 37 

⇒ n = 38

n = 38 is a positive integer, so yes, −100 is the 38th term.

Therefore,−100 is the 38th term of the AP.

Section B: 2-Mark Questions (Short Answer)

Q7: Find the 20th term from the last term of the AP: 3, 8, 13, …, 253.

Solution: Last term = 253, d = −5.

20th term from end = 253 + (20−1)(−5) = 253 − 95 = 158

20th term from last = 158

Q8: If the 3rd and 9th terms of an AP are 4 and −8 respectively, which term of the AP is zero?

Solution: a₃ = a + 2d = 4 and a₉ = a + 8d = −8. 

Subtracting: 6d = −12 

⇒ d = −2

a = 4 − 2(−2) = 8. 

Now aₙ = 0 

⇒ 8 + (n−1)(−2) = 0 

⇒ n−1 = 4 

⇒ n = 5

The 5th term of the AP is zero.

Q9: What is the common difference of an AP in which a₂₁ − a₇ = 84?

Solution: a₂₁ − a₇ = [a + 20d] − [a + 6d] = 14d = 84

d = 84/14 = 6

Common difference d = 6

Q10: In an AP, if d = −4, n = 7, and aₙ = 4, find the first term a.

Solution: aₙ = a + (n−1)d 

⇒ 4 = a + 6(−4) 

⇒ 4 = a − 24

a = 4 + 24 = 28

First term a = 28

Q11: Which term of the AP −7, −12, −17, −22, … will be −82? Is −100 a term of this AP?

Solution: a = −7, d = −5. aₙ = −82 

⇒ −7 + (n−1)(−5) = −82 

⇒ (n−1) = 75/5 = 15 

⇒ n = 16

For −100: −7 + (n−1)(−5) = −100 

⇒ (n−1) = 93/5 = 18.6, which is not an integer.

−82 is the 16th term. 

−100 is NOT a term of this AP (n is not a whole number).

Section C: 3-Mark Questions

Q12: How many terms of the AP 45, 39, 33, … must be taken so that their sum is 180?

Solution: a = 45, d = −6. 

Sₙ = 180 

⇒ n/2[2(45) + (n−1)(−6)] = 180

n/2 [90 − 6n + 6] = 180 

⇒ n(96 − 6n) = 360 

⇒ 6n² − 96n + 360 = 0 

⇒ n² − 16n + 60 = 0

⇒(n − 6)(n − 10) = 0 

⇒ n = 6 or n = 10

The AP is decreasing (d = −6). Terms after n = 6 include negative values (the 9th term = 45 + 8×(−6) = −3). These negative terms cancel out some of the positive sum, so both n = 6 and n = 10 give the same total of 180.

n = 6 or n = 10. Double answer because the AP contains negative terms that cancel positive contributions.

Q13: The sum of the first n terms of an AP is given by Sₙ = 3n² + 5n. Find the AP and its 25th term.

Solution: a₁ = S₁ = 3(1)² + 5(1) = 8

S₂ = 3(4) + 10 = 22 

⇒ a₂ = S₂ − S₁ = 22 − 8 = 14.

So d = 14 − 8 = 6

The AP is 8, 14, 20, 26, …

a₂₅ = 8 + 24(6) = 8 + 144 = 152

AP: 8, 14, 20, 26, …  

The 25th term = 152

Q14: The sum of 4 consecutive numbers in an AP is 32, and the product of the first and last terms to the product of the two middle terms is 7:15. Find the numbers.

Solution: Let 4 terms be a−3d, a−d, a+d, a+3d (common difference = 2d).

Sum = 4a = 32 

⇒ a = 8

Product ratio: (a−3d)(a+3d) : (a−d)(a+d) = 7 : 15 

⇒ (64−9d²)/(64−d²) = 7/15

⇒ 15(64−9d²) = 7(64−d²) 

⇒ 960 − 135d² = 448 − 7d² 

⇒ 128d² = 512 

⇒ d² = 4 

⇒ d = ±2

For d = 2: terms are 2, 6, 10, 14. 

For d = −2: terms are 14, 10, 6, 2 (same set).

The four numbers are 2, 6, 10, and 14.

Q15: If the ratio of the sum of the first n terms of two APs is (7n + 1) : (4n + 27), find the ratio of their 9th terms.

Solution: The ratio of nth terms of two APs = ratio of their sums formula with n replaced by (2n−1).

For 9th term: n = 9 

substitute (2×9 − 1) = 17 in the ratio expression.

Ratio = (7×17 + 1) : (4×17 + 27) = 120 : 95 = 24 : 19

Ratio of 9th terms = 24 : 19

Section D: Word Problems & Real Life Applications of Arithmetic Progression

Q17: A school is organising a charity run around a track. The first round is 300 m. Each subsequent round is 50 m longer. There are 10 rounds in total. Find the total distance covered in the event.

Solution: a = 300, d = 50, n = 10.

Sₙ = n/2[2a + (n−1)d] = 10/2[600 + 9×50] = 5[600 + 450] = 5 × 1050 = 5250 m

The 4th round: a₄ = 300 + 3(50) = 450 m  

5th round: 500 m  

6th round: 550 m

Total distance = 5250 metres. (4th, 5th, 6th rounds: 450 m, 500 m, 550 m)

Q18: In a potato race, a bucket is placed at the starting point 5 m from the first potato. Potatoes are placed 3 m apart. A competitor starts from the bucket, picks up the nearest potato, runs back to the bucket, picks up the next one, and so on. There are 10 potatoes. Find the total distance run. 

Solution: Distance for potato 1 = 2×5 = 10 m. 

Distance for potato 2 = 2×8 = 16 m. 

Distance for potato 3 = 2×11 = 22 m, …

This forms AP: 10, 16, 22, … with a = 10, d = 6, n = 10.

Sₙ = 10/2[2×10 + 9×6] 

= 5[20 + 54] 

= 5 × 74 = 370 m

Total distance run = 370 metres

Q19: Find the sum of all odd numbers between 100 and 200.

Solution: First odd number after 100 = 101, last odd number before 200 = 199. 

AP: 101, 103, …, 199. d = 2.

n = (199 − 101)/2 + 1 = 50. 

Sₙ = 50/2(101 + 199) = 25 × 300 = 7500

Sum = 7500

Section E: 4 & 5-Mark Questions with Case Studies

Q20 Case study: A school is organising a charity run to raise funds. The run consists of 10 rounds around a track. The first round is 300 m. Each subsequent round is 50 m longer than the previous one.

(i) Write the 4th, 5th, and 6th terms of the AP formed. (ii) Find the total distance covered in all 10 rounds. (iii) Which round will cover exactly 600 m?

Solution:

(i) AP: a = 300, d = 50. 

a₄ = 300 + 3(50) = 450 m 

a₅ = 500 m 

a₆ = 550 m

(ii) S₁₀ = 10/2[2(300) + 9(50)] = 5[600 + 450] = 5 × 1050 = 5250 m

(iii) aₙ = 600 

⇒ 300 + (n−1)(50) = 600 

⇒ (n−1) = 6 

⇒ n = 7

Q21: If m times the mth term of an AP equals n times the nth term, find the (m+n)th term.

Solution: m × aₘ = n × aₙ 

⇒ m[a + (m−1)d] = n[a + (n−1)d]

⇒ ma + m(m−1)d = na + n(n−1)d 

⇒ (m−n)a + [m²−m−n²+n]d = 0

⇒ (m−n)a + [(m−n)(m+n) − (m−n)]d = 0 

⇒ (m−n)[a + (m+n−1)d] = 0

Since m ≠ n: a + (m+n−1)d = 0 

⇒ This is exactly a(m+n) = 0

The (m+n)th term = 0

Q22: Find the sum of first 30 terms of an AP whose nth term is given by (3 + 2n).

Solution: Using AP formula: a₁ = 3+2(1) = 5, d = 2. 

S₃₀ = 30/2[2(5) + 29(2)]

= 15[10+58] 

= 15×68 = 1020 

Sum of first 30 terms = 1020

Q23: Find the sum of first 40 positive integers divisible by 3. Also find the sum of first 40 positive integers divisible by 6.

Solution: Multiples of 3: 3, 6, 9, …, 120.

a=3, d=3, n=40.

S₄₀ = 40/2[2(3)+39(3)] 

= 20[6+117] = 20×123 = 2460

Multiples of 6: 6, 12, 18, …, 240.

 a=6, d=6, n=40.

S₄₀ = 40/2[2(6)+39(6)] 

= 20[12+234] 

= 20×246 = 4920

Sum (multiples of 3) = 2460

Sum (multiples of 6) = 4920

Previous Year Questions on Arithmetic Progression

PYQ 2025: A charity run has 10 rounds; first round = 300 m, each subsequent round 50 m more. (i) Write 4th, 5th, 6th terms. (ii) Find the total distance. (iii) Which round = 600 m?

Solution: AP: 300, 350, 400, … 

a=300, d=50. 

(i) a₄=450, a₅=500, a₆=550. 

(ii) S₁₀ = 5(600+450) = 5250 m. 

(iii) aₙ=600 

⇒  n=7.

PYQ 2025: Three numbers in AP have sum 30. What is the middle term?

Solution: Let terms = a−d, a, a+d. 

Sum = 3a = 30 

⇒  a = 10.

Answer: Middle term = 10

PYQ 2024: The 8th term of an AP is 17 and its 14th term is 29. Find the AP.

Solution: a₈ = a+7d = 17 and a₁₄ = a+13d = 29.

Subtracting: 6d = 12 

⇒  d = 2. 

Then a = 17−14 = 3. AP: 3, 5, 7, 9, …

Answer: AP is 3, 5, 7, 9, … (a = 3, d = 2)

PYQ 2024: Find the sum of first 15 multiples of 8.

Solution: Multiples of 8: 8, 16, 24, … 

a=8, d=8, n=15. 

S₁₅ = 15/2[2(8)+14(8)] = 15/2 × 128 = 960.

Answer: Sum = 960

PYQ 2023: If the sum of first n terms of an AP is 4n − n², what is the first term? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, 10th, and nth terms.

Solution: S₁ = a₁ = 4(1)−1 = 3. 

S₂ = 4(2)−4 = 4 

⇒  a₂ = S₂−S₁ = 1. 

d = a₂−a₁ = −2.

aₙ = 3+(n−1)(−2) = 5−2n. 

a₃ = −1, a₁₀ = −15.

Therefore, a₁ = 3, a₂ = 1, a₃ = −1, a₁₀ = −15, aₙ = 5−2n

PYQ 2022: Find the middle term of the AP 6, 13, 20, …, 216. 

Solution: a=6, d=7. l=216 

⇒  6+(n−1)7=216 → n=31. 

Middle term = 16th term = 6+15×7 = 6+105 = 111.

Answer: Middle term = 111 (16th term of 31 terms)

PYQ 2021: The ratio of the sums of the first n terms of two APs is (7n + 1):(4n + 27). Find the ratio of their 9th terms.

Solution: Substitute n = 2(9)−1 = 17 in the ratio.

 (7×17+1):(4×17+27) = 120:95 = 24:19.

Answer: Ratio of 9th terms = 24 : 19

PYQ 2020: The sum of the first 30 terms of an AP is 1635. If its last term is 98, find the first term and common difference.

Solution: Sₙ = n/2(a+l)

1635 = 30/2(a+98) 

⇒  1635 = 15(a+98) 

⇒  a+98 = 109 

⇒  a = 11. 

d = (98−11)/29 = 87/29 = 3.

First term a = 11, Common difference d = 3