ClearStepsCLEARSTEPS AI
Teacher Login →Student Login →

Arithmetic Progressions: Class 10 Maths Practice Questions

30 original exam-pattern questions with full answers, matched to the current CBSE Class 10 paper design, including case-based questions. Attempt each question before opening the answer — or start a free 14-day trial ↓ for the full bank.

Q1Case-based4 marks

A school organises a 'Reading Marathon' over 30 days. On Day 1, a student reads 4 pages. Each day, the student reads 3 more pages than the previous day. The number of pages read each day forms an Arithmetic Progression with first term a = 4 and common difference d = 3.

A school organises a 'Reading Marathon' over 30 days. On Day 1, a student reads 4 pages. Each day, the student reads 3 more pages than the previous day.

(i) How many pages does the student read on Day 10? [1 mark]
(ii) On which day does the student read exactly 64 pages? [1 mark]
(iii) Find the total number of pages read by the student in the first 20 days. [2 marks]

OR

(iii) The student claims: "By the end of Day 15, I will have read more than 500 pages in total." Verify whether the student's claim is correct. [2 marks]

Show answer
This forms an A.P. with first term a = 4 and common difference d = 3.

(i) Pages read on Day 10:

Using aₙ = a + (n − 1)d

⟹ a₁₀ = 4 + (10 − 1) × 3

⟹ a₁₀ = 4 + 27

∴ The student reads 31 pages on Day 10.

(ii) Day on which student reads 64 pages:

Using aₙ = a + (n − 1)d

⟹ 64 = 4 + (n − 1) × 3

⟹ 60 = (n − 1) × 3

⟹ n − 1 = 20

⟹ n = 21

∴ The student reads exactly 64 pages on Day 21.

(iii) Total pages read in the first 20 days:

Using Sₙ = n/2 [2a + (n − 1)d]

⟹ S₂₀ = 20/2 [2 × 4 + (20 − 1) × 3]

⟹ S₂₀ = 10 [8 + 57]

⟹ S₂₀ = 10 × 65

∴ The total number of pages read in the first 20 days is 650 pages.

OR

(iii) Verifying the student's claim for the first 15 days:

Using Sₙ = n/2 [2a + (n − 1)d]

⟹ S₁₅ = 15/2 [2 × 4 + (15 − 1) × 3]

⟹ S₁₅ = 15/2 [8 + 42]

⟹ S₁₅ = 15/2 × 50

⟹ S₁₅ = 15 × 25

⟹ S₁₅ = 375

Since 375 < 500, the total pages read by the end of Day 15 is 375, which is NOT more than 500.

∴ The student's claim is incorrect. By the end of Day 15, the student will have read only 375 pages, which is less than 500 pages.
Q2Case-based4 marks

A school organises a 'Book Donation Drive' over several days. On the first day, students donate 8 books. Each subsequent day, they donate 6 more books than the previous day. The number of books donated each day forms an A.P. with first term a = 8 and common difference d = 6.

A school organises a 'Book Donation Drive' over several days. On the first day, students donate 8 books. Each subsequent day, they donate 6 more books than the previous day.

(i) How many books are donated on the 10th day? [1 mark]
(ii) What is the total number of books donated in the first 10 days? [1 mark]
(iii) On which day will the total books donated (from Day 1) first exceed 400? [2 marks]

Show answer
(i) The number of books donated each day forms an A.P. with a = 8 and d = 6.

Using aₙ = a + (n − 1)d,

a₁₀ = 8 + (10 − 1) × 6

⟹ a₁₀ = 8 + 54

∴ a₁₀ = 62 books.

---

(ii) Using Sₙ = n/2 [2a + (n − 1)d],

S₁₀ = 10/2 [2 × 8 + (10 − 1) × 6]

⟹ S₁₀ = 5 [16 + 54]

⟹ S₁₀ = 5 × 70

∴ S₁₀ = 350 books.

---

(iii) We need the smallest n such that Sₙ > 400.

Using Sₙ = n/2 [2a + (n − 1)d],

Sₙ = n/2 [2 × 8 + (n − 1) × 6]

⟹ Sₙ = n/2 [16 + 6n − 6]

⟹ Sₙ = n/2 [6n + 10]

⟹ Sₙ = n(3n + 5)

According to the question,

n(3n + 5) > 400

⟹ 3n² + 5n − 400 > 0

Solving 3n² + 5n − 400 = 0 using the quadratic formula:

n = [−5 ± √(25 + 4800)] / 6 = [−5 ± √4825] / 6

⟹ n = [−5 ± 69.46] / 6 (approximately)

Taking the positive root: n = (−5 + 69.46) / 6 ≈ 64.46 / 6 ≈ 10.74

Since n must be a whole number, the smallest integer value is n = 11.

Verification: S₁₁ = 11(3 × 11 + 5) = 11 × 38 = 418 > 400 ✓
S₁₀ = 350 < 400 ✓

∴ The total books donated first exceeds 400 on the 11th day.
Q3Case-based4 marks

A fitness trainer designs a running programme for an athlete. In the first week, the athlete runs 10 km. Each subsequent week, the athlete runs 2 km more than the previous week.

A fitness trainer designs a running programme for an athlete. In the first week, the athlete runs 10 km. Each subsequent week, the athlete runs 2 km more than the previous week.

(i) Write the A.P. formed by the weekly running distances and find the distance run in the 12th week. [1 mark]

(ii) In which week will the athlete first run more than 40 km? [1 mark]

(iii) The trainer sets a target: the athlete must complete a total of at least 500 km over the entire programme. Find the minimum number of weeks needed to meet this target.

OR

(iii) The athlete's friend follows a different programme where the total distance run after n weeks is given by S_n = 4n² + 6n km. Show that the friend's weekly running distances also form an A.P. and find its first term and common difference. [2 marks]

Show answer
(i) The weekly distances form an A.P. with first term a = 10 and common difference d = 2.

Using aₙ = a + (n − 1)d:

a₁₂ = 10 + (12 − 1) × 2

⟹ a₁₂ = 10 + 22

∴ The distance run in the 12th week = 32 km.

---

(ii) We need the first week n such that aₙ > 40.

aₙ = a + (n − 1)d

⟹ 10 + (n − 1) × 2 > 40

⟹ (n − 1) × 2 > 30

⟹ n − 1 > 15

⟹ n > 16

∴ The athlete will first run more than 40 km in the 17th week.

---

(iii) [Main Option]

Using Sₙ = n/2 [2a + (n − 1)d]:

Sₙ = n/2 [2(10) + (n − 1)(2)]

⟹ Sₙ = n/2 [20 + 2n − 2]

⟹ Sₙ = n/2 [2n + 18]

⟹ Sₙ = n(n + 9)

We need Sₙ ≥ 500:

n(n + 9) ≥ 500

⟹ n² + 9n − 500 ≥ 0

Using the quadratic formula with D = b² − 4ac = 81 + 2000 = 2081:

n = (−9 + √2081) / 2

√2081 ≈ 45.6

⟹ n ≥ (−9 + 45.6) / 2 ≈ 36.6 / 2 ≈ 18.3

∵ n must be a whole number and n ≥ 18.3,

∴ The minimum number of weeks needed = 19 weeks.

[Verification: S₁₉ = 19 × 28 = 532 ≥ 500 ✓; S₁₈ = 18 × 27 = 486 < 500 ✓]

---

(iii) [OR Option]

Given: Sₙ = 4n² + 6n

The distance run in the nth week = aₙ = Sₙ − Sₙ₋₁ (for n ≥ 2)

aₙ = [4n² + 6n] − [4(n − 1)² + 6(n − 1)]

⟹ aₙ = 4n² + 6n − 4(n² − 2n + 1) − 6n + 6

⟹ aₙ = 4n² + 6n − 4n² + 8n − 4 − 6n + 6

⟹ aₙ = 8n + 2

For n = 1: a₁ = S₁ = 4(1)² + 6(1) = 10 km.

Check using formula: a₁ = 8(1) + 2 = 10 ✓

Since aₙ = 8n + 2 is a linear function of n, the weekly distances form an A.P.

First term a = a₁ = 10 km.

Common difference d = aₙ − aₙ₋₁ = [8n + 2] − [8(n − 1) + 2] = 8.

∴ The friend's weekly distances form an A.P. with first term = 10 km and common difference = 8 km.
Q4Case-based4 marks

A construction company is stacking bricks in rows to form a triangular pattern. The bottom row has 45 bricks, and each successive row above it has 3 fewer bricks than the row below it. The top row has exactly 3 bricks.

A construction company is stacking bricks in rows to form a triangular pattern. The bottom row has 45 bricks, and each successive row above it has 3 fewer bricks than the row below it. The top row has exactly 3 bricks.

(i) How many rows of bricks are there in the stack? [1 mark]
(ii) Find the number of bricks in the 10th row from the bottom. [1 mark]
(iii) Find the total number of bricks in the entire stack. [2 marks]
OR
(iii) The company decides to use only those rows whose brick count is a multiple of 9. How many such rows exist, and what is the sum of bricks in those rows? [2 marks]

Show answer
The number of bricks in each row, counted from the bottom, forms an A.P.:
a = 45, d = −3, last term l = 3.

(i) Number of rows:

Using aₙ = a + (n − 1)d,

⟹ 3 = 45 + (n − 1)(−3)

⟹ 3 − 45 = (n − 1)(−3)

⟹ −42 = (n − 1)(−3)

⟹ n − 1 = 14

⟹ n = 15

∴ There are 15 rows of bricks in the stack.

(ii) Number of bricks in the 10th row from the bottom:

Using aₙ = a + (n − 1)d with n = 10, a = 45, d = −3,

⟹ a₁₀ = 45 + (10 − 1)(−3)

⟹ a₁₀ = 45 − 27

⟹ a₁₀ = 18

∴ The 10th row from the bottom has 18 bricks.

(iii) Total number of bricks in the entire stack:

Using Sₙ = n/2 (a + l) with n = 15, a = 45, l = 3,

⟹ S₁₅ = 15/2 × (45 + 3)

⟹ S₁₅ = 15/2 × 48

⟹ S₁₅ = 15 × 24

∴ The total number of bricks in the entire stack is 360.

OR

(iii) Rows whose brick count is a multiple of 9:

The brick counts form the A.P.: 45, 42, 39, 36, 33, 30, 27, 24, 21, 18, 15, 12, 9, 6, 3.

The terms that are multiples of 9 are: 45, 36, 27, 18, 9.

These form an A.P. with a = 45, d = −9.

Verifying the count using aₙ = a + (n − 1)d:

⟹ 9 = 45 + (n − 1)(−9)

⟹ −36 = (n − 1)(−9)

⟹ n − 1 = 4

⟹ n = 5

Sum of bricks in these 5 rows, using Sₙ = n/2 (a + l):

⟹ S₅ = 5/2 × (45 + 9)

⟹ S₅ = 5/2 × 54

⟹ S₅ = 5 × 27

∴ There are 5 such rows, and the total number of bricks in those rows is 135.
Q5Case-based4 marks

A school librarian arranges books on shelves in a systematic way. The first shelf has 8 books, the second shelf has 11 books, the third shelf has 14 books, and so on, following the same pattern.

A school librarian arranges books on shelves in a systematic way. The first shelf has 8 books, the second shelf has 11 books, the third shelf has 14 books, and so on, following the same pattern.

(i) Write the AP formed by the number of books on each shelf and find its common difference.
(ii) How many books are on the 15th shelf?
(iii) The librarian has a total of 200 books to place. Find the number of shelves needed to place all 200 books exactly. OR (iii) Which shelf contains 50 books?

Show answer
(i) The AP formed is: 8, 11, 14, 17, …

Here, a = 8

Common difference d = 11 − 8 = 3

∴ The AP is 8, 11, 14, 17, … with common difference d = 3.

(ii) Using the formula aₙ = a + (n − 1)d

a₁₅ = 8 + (15 − 1) × 3

⟹ a₁₅ = 8 + 14 × 3

⟹ a₁₅ = 8 + 42

∴ a₁₅ = 50 books

The 15th shelf contains 50 books.

(iii) Using the formula Sₙ = n/2 [2a + (n − 1)d]

According to the question, Sₙ = 200

n/2 [2 × 8 + (n − 1) × 3] = 200

⟹ n/2 [16 + 3n − 3] = 200

⟹ n/2 [3n + 13] = 200

⟹ n(3n + 13) = 400

⟹ 3n² + 13n − 400 = 0

Using the quadratic formula, D = b² − 4ac = (13)² − 4 × 3 × (−400) = 169 + 4800 = 4969

√4969 = √4969 ≈ 70.49, which is not a whole number.

∴ 200 books cannot be placed exactly on a whole number of shelves following this pattern.

OR

(iii) Let the nth shelf contain 50 books.

Using aₙ = a + (n − 1)d

50 = 8 + (n − 1) × 3

⟹ 50 − 8 = (n − 1) × 3

⟹ 42 = (n − 1) × 3

⟹ n − 1 = 14

⟹ n = 15

∴ The 15th shelf contains 50 books.
Q6Case-based4 marks

A fitness trainer designs a monthly workout challenge. In the first week, a participant completes 15 minutes of exercise per day. Each subsequent week, the daily exercise duration increases by 5 minutes.

A fitness trainer designs a monthly workout challenge. In the first week, a participant completes 15 minutes of exercise per day. Each subsequent week, the daily exercise duration increases by 5 minutes.

(i) Write the A.P. formed by the daily exercise durations (in minutes) for each week of the month (4 weeks). [1]
(ii) Find the total daily exercise duration accumulated over all 4 weeks (i.e., the sum of the A.P.). [1]
(iii) The trainer wants the participant to reach a daily exercise target of 95 minutes per day. In which week of the challenge will the participant first achieve this target? [2]

OR

(iii) The total exercise time (in minutes) over the first n weeks forms an A.P. whose sum is given by Sₙ = 5n² + 10n. Find the first term, the common difference, and the 10th term of the corresponding A.P. of weekly totals. [2]

Show answer
(i) The daily exercise durations (in minutes) for Weeks 1, 2, 3 and 4 are:
15, 20, 25, 30

Here, first term a = 15 and common difference d = 20 − 15 = 5.
∴ The A.P. is: 15, 20, 25, 30.

(ii) Using the formula Sₙ = n/2 [2a + (n−1)d] with n = 4, a = 15, d = 5:

S₄ = 4/2 [2(15) + (4−1)(5)]
⟹ S₄ = 2 [30 + 15]
⟹ S₄ = 2 × 45
∴ S₄ = 90 minutes.

(iii) [Main Option]
Using the general term formula aₙ = a + (n−1)d with a = 15, d = 5:

aₙ = 15 + (n−1)(5)
⟹ aₙ = 15 + 5n − 5
⟹ aₙ = 5n + 10

Setting aₙ = 95:
5n + 10 = 95
⟹ 5n = 85
⟹ n = 17

∴ The participant will first achieve the 95-minute daily target in Week 17 of the challenge.

(iii) [OR Option]
Given: Sₙ = 5n² + 10n

The nᵗʰ term of the A.P. of weekly totals is:
aₙ = Sₙ − Sₙ₋₁
⟹ aₙ = [5n² + 10n] − [5(n−1)² + 10(n−1)]
⟹ aₙ = [5n² + 10n] − [5(n² − 2n + 1) + 10n − 10]
⟹ aₙ = [5n² + 10n] − [5n² − 10n + 5 + 10n − 10]
⟹ aₙ = [5n² + 10n] − [5n² − 5]
⟹ aₙ = 10n + 5

First term (n = 1): a₁ = S₁ = 5(1)² + 10(1) = 5 + 10 = 15

Second term (n = 2): a₂ = 10(2) + 5 = 25

Common difference: d = a₂ − a₁ = 25 − 15 = 10

10th term: a₁₀ = 10(10) + 5 = 105

∴ First term = 15 minutes, Common difference = 10 minutes, and 10th term = 105 minutes.
Q7Case-based4 marks

A fitness trainer designs a running programme for an athlete. In the first week, the athlete runs 5 km. Each subsequent week, the athlete runs 2 km more than the previous week. The trainer sets a target: the athlete must complete a total of at least 120 km before stopping the programme.

A fitness trainer designs a running programme for an athlete. In the first week, the athlete runs 5 km. Each subsequent week, the athlete runs 2 km more than the previous week. The trainer sets a target: the athlete must complete a total of at least 120 km before stopping the programme.

(i) Write the A.P. formed by the weekly distances and state its first term and common difference. [1 mark]
(ii) Find the distance the athlete runs in the 10th week. [1 mark]
(iii) Find the minimum number of weeks required for the athlete to meet the 120 km target. [2 marks]

OR

(iii) The trainer later revises the programme. The athlete now runs distances each week that form an A.P. such that the 5th term is 21 km and the sum of the first 10 weeks is 200 km. Find the first term and the common difference of this revised A.P. [2 marks]

Show answer
(i) The weekly distances are: 5, 7, 9, 11, …
This forms an A.P. with first term a = 5 km and common difference d = 2 km.
∴ a = 5, d = 2.

(ii) Using aₙ = a + (n − 1)d,
⟹ a₁₀ = 5 + (10 − 1) × 2
⟹ a₁₀ = 5 + 18
∴ a₁₀ = 23 km.

(iii) Using Sₙ = n/2 [2a + (n − 1)d],
⟹ Sₙ = n/2 [2(5) + (n − 1)(2)]
⟹ Sₙ = n/2 [10 + 2n − 2]
⟹ Sₙ = n/2 [2n + 8]
⟹ Sₙ = n(n + 4)

We require Sₙ ≥ 120.
⟹ n(n + 4) ≥ 120
⟹ n² + 4n − 120 ≥ 0
⟹ (n + 14)(n − 10) ≥ 0

∵ n must be a positive integer, we reject n = −14.
⟹ n ≥ 10.

∴ The minimum number of weeks required is 10.

Verification: S₁₀ = 10 × 14 = 140 ≥ 120 ✓ and S₉ = 9 × 13 = 117 < 120 ✓

OR

(iii) Let the first term of the revised A.P. be a and common difference be d.

Using aₙ = a + (n − 1)d:
a₅ = a + 4d = 21 …(i)

Using Sₙ = n/2 [2a + (n − 1)d]:
S₁₀ = 10/2 [2a + 9d] = 200
⟹ 5(2a + 9d) = 200
⟹ 2a + 9d = 40 …(ii)

From (i): a = 21 − 4d …(iii)

Substituting (iii) into (ii):
⟹ 2(21 − 4d) + 9d = 40
⟹ 42 − 8d + 9d = 40
⟹ d = −2

Substituting d = −2 into (iii):
⟹ a = 21 − 4(−2) = 21 + 8 = 29

∴ The first term a = 29 km and the common difference d = −2 km.
Q8Short Answer1 mark

Assertion (A): In an AP, the nth term is a + (n-1)d.
Reason (R): An AP is a sequence where consecutive terms have a constant difference.

Show answer
Option (a) is correct.

Explanation: Assertion (A) is true: the nth term of an AP with first term a and common difference d is aₙ = a + (n−1)d. Reason (R) is also true: an AP is defined as a sequence in which each term differs from the preceding term by a constant value d. R is the correct explanation of A because the formula aₙ = a + (n−1)d follows directly from this definition — starting at a and adding d exactly (n−1) times yields a + (n−1)d.
Q9Short Answer1 mark

Assertion (A): In an AP, the nth term is a + (n-1)d.
Reason (R): An AP is a sequence where consecutive terms have a constant difference.

Show answer
Option (a) is correct.

Explanation: R states that an AP is a sequence where consecutive terms differ by a constant d — this is the definition of an AP. A states the nth term formula aₙ = a + (n−1)d, which is derived directly from R: starting at the first term a and adding d exactly (n−1) times. Both A and R are true, and R is the correct explanation of A.
Q10Short Answer1 mark

Assertion (A): In an AP, the nth term is a + (n-1)d.
Reason (R): An AP is a sequence where consecutive terms have a constant difference.

Show answer
Option (a) is correct.

Explanation: Assertion (A) is true: the n<super>th</super> term of an AP is a<sub>n</sub> = a + (n−1)d, which is the standard result. Reason (R) is true: an AP is defined as a sequence in which the difference between consecutive terms is constant (the common difference d). Since a<sub>n</sub> = a + (n−1)d is derived directly from the definition stated in R — each successive term adds one more d to the first term — R is the correct explanation of A.
Q11Short Answer1 mark

Assertion (A): In an AP, the nth term is a + (n-1)d.
Reason (R): An AP is a sequence where consecutive terms have a constant difference.

Show answer
Option (a) is correct.

Explanation: Assertion A states that the n<super>th</super> term of an AP is aₙ = a + (n−1)d — this is the standard formula and is TRUE. Reason R states that an AP is a sequence where consecutive terms have a constant difference d — this definition is also TRUE. Since aₙ = a + (n−1)d is derived directly from the property of constant common difference (a + 0d, a + 1d, a + 2d, … giving the n<super>th</super> term as a + (n−1)d), R is the correct explanation of A.
Q12Short Answer1 mark

Assertion (A): In an AP, the nth term is a + (n-1)d.
Reason (R): An AP is a sequence where consecutive terms have a constant difference.

Show answer
Option (a) is correct.

Explanation: Assertion (A) states that the n<super>th</super> term of an AP is aₙ = a + (n − 1)d, which is the standard formula for the general term and is true. Reason (R) states that an AP is a sequence where consecutive terms have a constant difference d, which is the definition of an AP and is also true. Since the formula aₙ = a + (n − 1)d is derived directly from this definition — beginning at first term a and adding d exactly (n − 1) times — R is the correct explanation of A.
Q13Short Answer1 mark

Assertion (A): In an AP, the nth term is a + (n-1)d.
Reason (R): An AP is a sequence where consecutive terms have a constant difference.

Show answer
Option (a) is correct.

Explanation: Assertion A is true — the n<super>th</super> term of an AP is a<sub>n</sub> = a + (n−1)d, obtained by adding the common difference d exactly (n−1) times to the first term a. Reason R is true — an AP is defined as a sequence in which each consecutive pair of terms has a constant difference d. R is the correct explanation of A because it is precisely this constant-difference property that gives rise to the formula a<sub>n</sub> = a + (n−1)d.
Q14Short Answer3 marks

The term of an A.P. is given by . Find the sum of the first 24 terms of this A.P.

Show answer
Given: aₙ = 3n + 2

Finding a₁ and a₂₄:

a₁ = 3(1) + 2 = 5

a₂₄ = 3(24) + 2 = 72 + 2 = 74

Finding the common difference:

a₂ = 3(2) + 2 = 8

⟹ d = a₂ − a₁ = 8 − 5 = 3

This confirms an A.P. with a = 5 and d = 3.

Finding S₂₄:

Using Sₙ = n/2 (a + l), where l = a₂₄ = 74:

S₂₄ = 24/2 × (a₁ + a₂₄)

⟹ S₂₄ = 12 × (5 + 74)

⟹ S₂₄ = 12 × 79

∴ S₂₄ = 948
Q15Short Answer3 marks

The sum of the first 6 terms of an A.P. is equal to the sum of its first 14 terms. Show that the sum of its first 20 terms is zero.

Show answer
Let the first term be a and common difference be d.

S<sub>n</sub> = n/2[2a + (n − 1)d]

Given: S<sub>6</sub> = S<sub>14</sub>

⟹ 6/2[2a + 5d] = 14/2[2a + 13d]

⟹ 3[2a + 5d] = 7[2a + 13d]

⟹ 6a + 15d = 14a + 91d

⟹ 0 = 8a + 76d

⟹ 8a + 76d = 0

⟹ 2a + 19d = 0 &nbsp;&nbsp;&nbsp;&nbsp;...(i)

To show: S<sub>20</sub> = 0

S<sub>20</sub> = 20/2[2a + 19d]

⟹ S<sub>20</sub> = 10[2a + 19d]

From (i), 2a + 19d = 0

⟹ S<sub>20</sub> = 10 × 0 = 0

∴ The sum of the first 20 terms of the A.P. is zero. Hence proved.
Q16Short Answer3 marks

If the sum of the first p terms of an A.P. is equal to the sum of its first q terms (p ≠ q), show that the sum of its first (p + q) terms is zero.

Show answer
Let the first term of the A.P. be *a* and the common difference be *d*.

The sum of the first *n* terms is given by:

S<sub>n</sub> = n/2 · [2a + (n − 1)d]

Given: S<sub>p</sub> = S<sub>q</sub>, where p ≠ q.

⟹ p/2 · [2a + (p − 1)d] = q/2 · [2a + (q − 1)d]

⟹ p[2a + (p − 1)d] = q[2a + (q − 1)d]

⟹ 2ap + p(p − 1)d = 2aq + q(q − 1)d

⟹ 2a(p − q) + d[p(p − 1) − q(q − 1)] = 0

⟹ 2a(p − q) + d[p² − p − q² + q] = 0

⟹ 2a(p − q) + d[(p² − q²) − (p − q)] = 0

⟹ 2a(p − q) + d(p − q)(p + q − 1) = 0

Since p ≠ q, dividing throughout by (p − q):

⟹ 2a + d(p + q − 1) = 0 &nbsp;&nbsp;&nbsp;&nbsp; …(i)

To show: S<sub>p+q</sub> = 0.

S<sub>p+q</sub> = (p + q)/2 · [2a + (p + q − 1)d]

⟹ S<sub>p+q</sub> = (p + q)/2 × 0 &nbsp;&nbsp;&nbsp;&nbsp; [using (i)]

∴ S<sub>p+q</sub> = 0.

Hence, the sum of the first (p + q) terms of the A.P. is zero. Hence proved.
Q17Short Answer3 marks

The term of an A.P. is given by . Find the sum of the first 18 terms of this A.P.

Show answer
Given: aₙ = 4n − 1

Finding the first term and common difference:

a₁ = 4(1) − 1 = 3

a₂ = 4(2) − 1 = 7

⟹ d = a₂ − a₁ = 7 − 3 = 4

This forms an A.P. with a = 3, d = 4.

Finding the sum of the first 18 terms:

Using the formula Sₙ = n/2 [2a + (n − 1)d],

S₁₈ = 18/2 [2(3) + (18 − 1)(4)]

⟹ S₁₈ = 9 [6 + 68]

⟹ S₁₈ = 9 × 74

∴ S₁₈ = 666
Q18Short Answer3 marks

In an A.P., the sum of five consecutive terms is 55 and the sum of their squares is 645. Find the five terms of the A.P.

Show answer
Let the five consecutive terms of the A.P. be a − 2d, a − d, a, a + d, a + 2d.

Sum of five terms:

(a − 2d) + (a − d) + a + (a + d) + (a + 2d) = 55

⟹ 5a = 55

⟹ a = 11 &emsp;...(i)

Sum of squares of five terms:

(a − 2d)² + (a − d)² + a² + (a + d)² + (a + 2d)² = 645

Expanding and collecting like terms:

⟹ 5a² + 10d² = 645 &emsp;...(ii)

Substituting a = 11 from (i) into (ii):

⟹ 5(11)² + 10d² = 645

⟹ 5(121) + 10d² = 645

⟹ 605 + 10d² = 645

⟹ 10d² = 40

⟹ d² = 4

⟹ d = ±2

Finding the five terms:

When d = 2: the five terms are 7, 9, 11, 13, 15.

When d = −2: the five terms are 15, 13, 11, 9, 7 (same set in reverse order).

∴ The five terms of the A.P. are 7, 9, 11, 13, 15.
Q19Short Answer3 marks

In an A.P., the sum of three consecutive terms is 27 and the sum of their squares is 293. Find the three terms of the A.P.

Show answer
Let the three consecutive terms of the A.P. be (a − d), a, (a + d).

Condition 1: Sum of the three terms = 27

(a − d) + a + (a + d) = 27

⟹ 3a = 27

⟹ a = 9

Condition 2: Sum of squares of the three terms = 293

(a − d)<super>2</super> + a<super>2</super> + (a + d)<super>2</super> = 293

⟹ a<super>2</super> − 2ad + d<super>2</super> + a<super>2</super> + a<super>2</super> + 2ad + d<super>2</super> = 293

⟹ 3a<super>2</super> + 2d<super>2</super> = 293

Substituting a = 9:

⟹ 3(9)<super>2</super> + 2d<super>2</super> = 293

⟹ 3(81) + 2d<super>2</super> = 293

⟹ 243 + 2d<super>2</super> = 293

⟹ 2d<super>2</super> = 50

⟹ d<super>2</super> = 25

⟹ d = ±5

When d = 5: the three terms are (9 − 5), 9, (9 + 5) = 4, 9, 14.

When d = −5: the three terms are (9 + 5), 9, (9 − 5) = 14, 9, 4.

∴ The three terms of the A.P. are 4, 9, 14 (or 14, 9, 4 in reverse order).
Q20Short Answer3 marks

Scheme A offers a savings plan where the monthly deposit starts at ₹500 in the first month and increases by ₹50 each subsequent month. Scheme B offers a savings plan where the monthly deposit starts at ₹200 in the first month and increases by ₹100 each subsequent month. Find the total amount deposited under each scheme over 12 months and determine which scheme results in a higher total deposit.

Show answer
Scheme A: Monthly deposits form an A.P. with a = 500, d = 50, n = 12.

Using S<sub>n</sub> = n/2 [2a + (n − 1)d],

S<sub>12</sub> = 12/2 [2(500) + (12 − 1)(50)]

⟹ S<sub>12</sub> = 6 [1000 + 550]

⟹ S<sub>12</sub> = 6 × 1550

∴ Total deposit under Scheme A = ₹9,300

---

Scheme B: Monthly deposits form an A.P. with a = 200, d = 100, n = 12.

Using S<sub>n</sub> = n/2 [2a + (n − 1)d],

S<sub>12</sub> = 12/2 [2(200) + (12 − 1)(100)]

⟹ S<sub>12</sub> = 6 [400 + 1100]

⟹ S<sub>12</sub> = 6 × 1500

∴ Total deposit under Scheme B = ₹9,000

---

Comparison:

Since ₹9,300 > ₹9,000,

∴ Scheme A results in a higher total deposit of ₹9,300 over 12 months, which is ₹300 more than the total deposit of ₹9,000 under Scheme B.
Q21Short Answer3 marks

The ratio of the 7th term to the 14th term of an A.P. is 1 : 2, and the sum of its first 10 terms is 115. Find the first term and the common difference of the A.P.

Show answer
Let the first term be *a* and the common difference be *d*.

aₙ = a + (n − 1)d

Condition 1: a₇/a₁₄ = 1/2

⟹ (a + 6d)/(a + 13d) = 1/2

⟹ 2(a + 6d) = a + 13d

⟹ 2a + 12d = a + 13d

⟹ a = d …(i)

Condition 2: S₁₀ = 115

Using Sₙ = n/2 [2a + (n − 1)d]:

S₁₀ = 10/2 [2a + 9d] = 5(2a + 9d) = 115

⟹ 2a + 9d = 23 …(ii)

Solving (i) and (ii):

Substituting a = d in (ii):

⟹ 2d + 9d = 23

⟹ 11d = 23

⟹ d = 23/11

From (i): a = d = 23/11

Verification: S₁₀ = 5(2 × 23/11 + 9 × 23/11) = 5 × (23/11)(2 + 9) = 5 × (23/11) × 11 = 5 × 23 = 115 ✓

∴ The first term a = 23/11 and the common difference d = 23/11.
Q22Short Answer3 marks

Scheme A offers a fixed annual increment of ₹1,200 on a starting salary of ₹18,000. Scheme B offers a fixed annual increment of ₹900 on a starting salary of ₹20,000. Determine the total salary earned under each scheme over 10 years and state which scheme gives a higher total earning.

Show answer
Scheme A: Starting salary a = ₹18,000, annual increment d = ₹1,200, n = 10.

The yearly salaries form an A.P. with a = 18000, d = 1200.

Using Sₙ = n/2 [2a + (n − 1)d]:

S_A = 10/2 [2 × 18000 + (10 − 1) × 1200]

⟹ S_A = 5 [36000 + 10800]

⟹ S_A = 5 × 46800

∴ S_A = ₹2,34,000

---

Scheme B: Starting salary a = ₹20,000, annual increment d = ₹900, n = 10.

The yearly salaries form an A.P. with a = 20000, d = 900.

Using Sₙ = n/2 [2a + (n − 1)d]:

S_B = 10/2 [2 × 20000 + (10 − 1) × 900]

⟹ S_B = 5 [40000 + 8100]

⟹ S_B = 5 × 48100

∴ S_B = ₹2,40,500

---

Since S_B = ₹2,40,500 > S_A = ₹2,34,000,

∴ Scheme B gives a higher total earning over 10 years by ₹6,500.
Q23Short Answer3 marks

AP Series A has first term 3 and common difference 4. AP Series B has first term 7 and common difference 2. The sum of the first n terms of Series A equals the sum of the first n terms of Series B for some value of n > 0. Find that value of n. Also determine which series has a greater 10th term and by how much.

Show answer
For Series A: a = 3, d = 4.

Using Sₙ = n/2[2a + (n−1)d],

S_A = n/2[2(3) + (n−1)(4)] = n/2[6 + 4n − 4] = n/2(4n + 2) = n(2n + 1)

For Series B: a = 7, d = 2.

S_B = n/2[2(7) + (n−1)(2)] = n/2[14 + 2n − 2] = n/2(2n + 12) = n(n + 6)

Setting S_A = S_B:

n(2n + 1) = n(n + 6)

Since n > 0, dividing both sides by n:

2n + 1 = n + 6

⟹ n = 5

∴ The sum of the first 5 terms of Series A equals the sum of the first 5 terms of Series B.

Using aₙ = a + (n−1)d:

10th term of Series A: T₁₀ = 3 + (10 − 1)(4) = 3 + 36 = 39

10th term of Series B: T₁₀ = 7 + (10 − 1)(2) = 7 + 18 = 25

Difference = 39 − 25 = 14

∴ The 10th term of Series A is greater than the 10th term of Series B by 14.
Q24Short Answer3 marks

The term of an A.P. is given by . Find the sum of the first 15 terms of this A.P. Also, determine whether 87 is a term of this A.P.

Show answer
Given: aₙ = 3 + 4n

First term and common difference:

a₁ = 3 + 4(1) = 7

a₂ = 3 + 4(2) = 11

∴ d = a₂ − a₁ = 11 − 7 = 4

This forms an A.P. with a = 7, d = 4.

Sum of first 15 terms:

Using Sₙ = n/2 [2a + (n − 1)d],

S₁₅ = 15/2 [2(7) + (15 − 1)(4)]

⟹ S₁₅ = 15/2 [14 + 56]

⟹ S₁₅ = 15/2 × 70

∴ S₁₅ = 525

Whether 87 is a term of the A.P.:

Let aₙ = 87.

⟹ 3 + 4n = 87

⟹ 4n = 84

⟹ n = 21

Since n = 21 is a positive integer, ∴ 87 is the 21st term of the A.P.
Q25Short Answer3 marks

If the sum of the first 10 terms of an A.P. is 100 and the sum of its first 30 terms is 900, find the sum of its first 40 terms.

Show answer
Let the first term be a and common difference be d.

S<sub>n</sub> = n/2 [2a + (n−1)d]

S<sub>10</sub> = 10/2 [2a + 9d] = 100

⟹ 5(2a + 9d) = 100

⟹ 2a + 9d = 20 ...(i)

S<sub>30</sub> = 30/2 [2a + 29d] = 900

⟹ 15(2a + 29d) = 900

⟹ 2a + 29d = 60 ...(ii)

Subtracting (i) from (ii):

20d = 40

⟹ d = 2

Substituting d = 2 in (i):

2a + 9(2) = 20

⟹ 2a = 20 − 18 = 2

⟹ a = 1

S<sub>40</sub> = 40/2 [2(1) + 39(2)]

= 20 [2 + 78]

= 20 × 80

∴ S<sub>40</sub> = 1600
Q26Short Answer3 marks

In an A.P., the sum of three consecutive terms is 33 and the product of the first and third terms is 105. Find the three terms of the A.P.

Show answer
Let the three consecutive terms of the A.P. be (a − d), a, (a + d).

Sum condition:

(a − d) + a + (a + d) = 33

⟹ 3a = 33

⟹ a = 11

Product condition:

(a − d)(a + d) = 105

⟹ a² − d² = 105

⟹ (11)² − d² = 105

⟹ 121 − d² = 105

⟹ d² = 16

⟹ d = ±4

When d = 4: the three terms are 7, 11, 15.

When d = −4: the three terms are 15, 11, 7 (same terms in reverse order).

∴ The three terms of the A.P. are 7, 11, 15.
Q27Short Answer3 marks

Find the A.P. whose fifth term is 26 and ninth term exceeds the seventh term by 8. Also, find the sum of first 35 terms of the A.P.

Show answer
Let the first term be a and the common difference be d.

Using aₙ = a + (n − 1)d:

a₅ = a + 4d = 26 …(i)

Ninth term exceeds seventh term by 8:

a₉ − a₇ = 8

⟹ (a + 8d) − (a + 6d) = 8

⟹ 2d = 8

⟹ d = 4

Substituting d = 4 in (i):

a + 4(4) = 26

⟹ a = 26 − 16 = 10

∴ The A.P. is: 10, 14, 18, 22, …

Sum of first 35 terms, using Sₙ = n/2 [2a + (n − 1)d]:

S₃₅ = 35/2 [2(10) + (35 − 1)(4)]

⟹ S₃₅ = 35/2 [20 + 136]

⟹ S₃₅ = 35/2 × 156

⟹ S₃₅ = 35 × 78

∴ S₃₅ = 2730
Q28Short Answer3 marks

Series A is an AP with first term 5 and common difference 4. Series B is an AP with first term 20 and common difference 1. Determine which series has a greater sum of first 15 terms and by how much.

Show answer
For an AP, Sₙ = n/2[2a + (n−1)d].

Series A: a = 5, d = 4, n = 15

S₁₅ᴬ = 15/2[2(5) + (14)(4)]

⟹ = 15/2[10 + 56]

⟹ = 15/2 × 66

⟹ = 15 × 33 = 495

Series B: a = 20, d = 1, n = 15

S₁₅ᴮ = 15/2[2(20) + (14)(1)]

⟹ = 15/2[40 + 14]

⟹ = 15/2 × 54

⟹ = 15 × 27 = 405

Since 495 > 405,

Difference = 495 − 405 = 90

∴ Series A has a greater sum of first 15 terms, and it exceeds the sum of Series B by 90.
Q29Short Answer3 marks

For an A.P., it is given that the first term (a) = 7, common difference (d) = 4, and the nth term (aₙ) = 79. Find n and the sum of first n terms (Sₙ) of the A.P.

Show answer
Given: a = 7, d = 4, aₙ = 79.

Using aₙ = a + (n − 1)d:

79 = 7 + (n − 1) × 4

⟹ 72 = (n − 1) × 4

⟹ n − 1 = 18

⟹ n = 19

Using Sₙ = n/2 (a + aₙ):

S₁₉ = 19/2 × (7 + 79)

⟹ S₁₉ = 19/2 × 86

⟹ S₁₉ = 19 × 43

∴ n = 19 and S₁₉ = 817.
Q30Short Answer3 marks

Show that the sum of all terms of an A.P. whose first term is , second term is and last term is , is equal to .

Show answer
Let the first term a = p, common difference = d, and number of terms = n, with last term = r.

Since the second term is q:

q = p + d ⟹ d = q − p

Using the last term formula aₙ = a + (n − 1)d:

r = p + (n − 1)(q − p)

⟹ (n − 1)(q − p) = r − p

⟹ n − 1 = (r − p)/(q − p)

⟹ n = (r − p)/(q − p) + 1 = (r − p + q − p)/(q − p) = (q + r − 2p)/(q − p)

Using the sum formula Sₙ = n/2 (first term + last term):

S = n/2 · (p + r)

Substituting n = (q + r − 2p)/(q − p):

S = 1/2 · [(q + r − 2p)/(q − p)] · (p + r)

∴ S = (p + r)(q + r − 2p) / 2(q − p)

Hence proved.

Want unlimited practice on Arithmetic Progressions?

The full ClearSteps bank has 43+ reviewed questions on this chapter alone — with step-wise solutions, difficulty levels, and progress tracking. Verify your WhatsApp number and your free trial starts right here.

CBSE · CLASS 10
🎟️14-day free trial · code PRAC auto-applied
Mobile Number
+91
OTP will be sent to your WhatsApp

⭐ Trusted by 12,000+ students across India

All chapters →Chapter-wise PYQ →Sample Papers →
Arithmetic Progressions — Class 10 Maths Practice Questions