Arithmetic Progressions Test Class 10 Maths — 40 Marks
Pandey Sir’s BEST Classes
PERIODIC TEST — Arithmetic Progressions (2026-27) | CLASS 10 | MATHEMATICS
Max Marks: 40 · Time: 1 hour 30 minutes (Time left: 90:00)
| Section | Marks | Questions | Time | Strategy |
|---|---|---|---|---|
| A | 12 | 12 MCQ | 15 min | Quick recall — about 1 minute each |
| B | 8 | 4 Short | 20 min | Show the formula before substituting |
| C | 12 | 4 Long | 30 min | Full steps, units, final answer boxed |
| D | 8 | 2 Case | 20 min | Read the paragraph, label the parts |
| Review | — | — | 5 min | Cross-check totals and answers |
General Instructions: All questions are compulsory. Section A has 12 MCQs of 1 mark each. Internal choice is given in Sections B, C and D — attempt any ONE of the two options. Use π = 22/7 only if a question says so. Draw diagrams only where asked.
Section A — Multiple Choice (12 × 1 = 12 marks)
Choose the single best option.
The common difference of the AP 5, 8, 11, 14, … is
Marking Scheme: Correct option: 1 (no partial marks).
Total: 1 Mark
Total: 1 Mark
Common difference
d = a2 − a1
= 8 − 5
= 3
= 8 − 5
= 3
Answer: (b) 3
Which of the following lists of numbers forms an arithmetic progression?
Marking Scheme: Correct option: 1 (no partial marks).
Total: 1 Mark
Total: 1 Mark
Test for a common difference
(a) 4−2=2 but 8−4=4 → not constant
(b) differences 3, 5, 7 → not constant
(c) 3−3=0, 3−3=0 → constant, d = 0
(d) differences 1, 2, 3 → not constant
(b) differences 3, 5, 7 → not constant
(c) 3−3=0, 3−3=0 → constant, d = 0
(d) differences 1, 2, 3 → not constant
Answer: (c) 3, 3, 3, 3 — a constant sequence is an AP with d = 0
The 10th term of the AP 2, 7, 12, … is
Marking Scheme: Correct option: 1 (no partial marks).
Total: 1 Mark
Total: 1 Mark
nth term
a = 2, d = 5
an = a + (n − 1)d
a10 = 2 + (10 − 1)×5
= 2 + 45
= 47
an = a + (n − 1)d
a10 = 2 + (10 − 1)×5
= 2 + 45
= 47
Answer: (a) 47
If a = −3 and d = 4, then the first four terms of the AP are
Marking Scheme: Correct option: 1 (no partial marks).
Total: 1 Mark
Total: 1 Mark
Add the common difference each time
a1 = −3
a2 = −3 + 4 = 1
a3 = 1 + 4 = 5
a4 = 5 + 4 = 9
a2 = −3 + 4 = 1
a3 = 1 + 4 = 5
a4 = 5 + 4 = 9
Answer: (a) −3, 1, 5, 9
The sum of the first 20 natural numbers is
Marking Scheme: Correct option: 1 (no partial marks).
Total: 1 Mark
Total: 1 Mark
Sum of first n natural numbers
Sn = n(n + 1)2
S20 = 20 × 212
= 10 × 21
= 210
S20 = 20 × 212
= 10 × 21
= 210
Answer: (b) 210
For an AP whose nth term is an = 3n + 2, the first term is
Marking Scheme: Correct option: 1 (no partial marks).
Total: 1 Mark
Total: 1 Mark
Put n = 1
a1 = 3(1) + 2
= 5
= 5
Answer: (c) 5
If the sum of the first n terms of an AP is Sn = n2 + n, then its second term is
Marking Scheme: Correct option: 1 (no partial marks).
Total: 1 Mark
Total: 1 Mark
Term from sums
an = Sn − Sn−1
S1 = 1 + 1 = 2, so a1 = 2
S2 = 4 + 2 = 6
a2 = S2 − S1 = 6 − 2 = 4
S1 = 1 + 1 = 2, so a1 = 2
S2 = 4 + 2 = 6
a2 = S2 − S1 = 6 − 2 = 4
Answer: (b) 4
The 11th term of the AP −5, −52, 0, 52, … is
Marking Scheme: Correct option: 1 (no partial marks).
Total: 1 Mark
Total: 1 Mark
Common difference
d = a2 − a1 = −5/2 − (−5)
= −5/2 + 5 = 5/2
= −5/2 + 5 = 5/2
11th term
a11 = a + 10d
= −5 + 10 × 5/2
= −5 + 25
= 20
= −5 + 10 × 5/2
= −5 + 25
= 20
Answer: (b) 20
Which term of the AP 21, 18, 15, … is −81?
Marking Scheme: Correct option: 1 (no partial marks).
Total: 1 Mark
Total: 1 Mark
Set the nth term equal to −81
a = 21, d = −3
a + (n − 1)d = −81
21 + (n − 1)(−3) = −81
(n − 1)(−3) = −102
n − 1 = 34
n = 35
a + (n − 1)d = −81
21 + (n − 1)(−3) = −81
(n − 1)(−3) = −102
n − 1 = 34
n = 35
Answer: (b) 35th term
If the 13th term of an AP exceeds its 7th term by 12, then the common difference is
Marking Scheme: Correct option: 1 (no partial marks).
Total: 1 Mark
Total: 1 Mark
Difference of two terms
a13 − a7 = 12
(a + 12d) − (a + 6d) = 12
6d = 12
d = 2
(a + 12d) − (a + 6d) = 12
6d = 12
d = 2
Answer: (a) 2
The sum of the first n odd natural numbers is
Marking Scheme: Correct option: 1 (no partial marks).
Total: 1 Mark
Total: 1 Mark
Use the sum formula
Odd numbers: 1, 3, 5, … with a = 1, d = 2
Sn = n/2 [2(1) + (n − 1)2]
= n/2 [2 + 2n − 2]
= n/2 × 2n
= n2
Sn = n/2 [2(1) + (n − 1)2]
= n/2 [2 + 2n − 2]
= n/2 × 2n
= n2
Answer: (a) n2
In an AP, if a = 5 and d = 0, then the 15th term is
Marking Scheme: Correct option: 1 (no partial marks).
Total: 1 Mark
Total: 1 Mark
Constant sequence
a15 = a + 14d
= 5 + 14 × 0
= 5
= 5 + 14 × 0
= 5
Answer: (c) 5 — every term equals the first term when d = 0
Section B — Short Answer (4 × 2 = 8 marks)
Find the 15th term of the AP 7, 11, 15, …
Marking Scheme: Formula 1 mark · Substitution and answer 1 mark.
Total: 2 Marks
Total: 2 Marks
Identify a and d
a = 7
d = 11 − 7 = 4
d = 11 − 7 = 4
Apply the nth-term formula
an = a + (n − 1)d
a15 = 7 + 14 × 4
= 7 + 56
= 63
a15 = 7 + 14 × 4
= 7 + 56
= 63
Answer: a15 = 63
How many two-digit numbers are divisible by 3?
Marking Scheme: Correct AP identified 1 mark · Solving for n 1 mark.
Total: 2 Marks
Total: 2 Marks
Write the AP
Two-digit multiples of 3: 12, 15, 18, …, 99
a = 12, d = 3, last term = 99
a = 12, d = 3, last term = 99
Solve for the number of terms
12 + (n − 1)3 = 99
(n − 1)3 = 87
n − 1 = 29
n = 30
(n − 1)3 = 87
n − 1 = 29
n = 30
Answer: 30 two-digit numbers
The first term of an AP is 5, the last term is 45 and the sum of all its terms is 400. Find the number of terms.
Marking Scheme: Using the sum formula 1 mark · Solving and answer 1 mark.
Total: 2 Marks
Total: 2 Marks
Sum formula with the last term
Sn = n2[a + l]
400 = n2[5 + 45]
400 = n2 × 50
400 = n2[5 + 45]
400 = n2 × 50
Solve for n
400 = 25n
n = 16
n = 16
Answer: 16 terms
If the sum of the first n terms of an AP is given by Sn = 3n2 + 5n, find its 10th term.
Option 1 — Sn = 3n2 + 5n
Marking Scheme: Using an = Sn − Sn−1 1 mark · Answer 1 mark.
Total: 2 Marks
Total: 2 Marks
Subtract consecutive sums
an = Sn − Sn−1
= (3n2 + 5n) − [3(n − 1)2 + 5(n − 1)]
= (3n2 + 5n) − (3n2 − 6n + 3 + 5n − 5)
= (3n2 + 5n) − (3n2 − n − 2)
= 6n + 2
= (3n2 + 5n) − [3(n − 1)2 + 5(n − 1)]
= (3n2 + 5n) − (3n2 − 6n + 3 + 5n − 5)
= (3n2 + 5n) − (3n2 − n − 2)
= 6n + 2
Put n = 10
a10 = 6(10) + 2
= 62
= 62
Answer: a10 = 62
— OR —
Option 2
Find the 20th term from the last term of the AP 3, 8, 13, …, 253.
Marking Scheme: Identifying the reversed AP 1 mark · Answer 1 mark.
Total: 2 Marks
Total: 2 Marks
Reverse the AP
Taking the last term as the first:
a = 253, d = −5
a = 253, d = −5
20th term
a20 = 253 + 19(−5)
= 253 − 95
= 158
= 253 − 95
= 158
Answer: 158
[Attempt any ONE of the two options of Q16]
Section C — Long Answer (4 × 3 = 12 marks)
The sum of the 4th and 8th terms of an AP is 24, and the sum of its 6th and 10th terms is 44. Find the first three terms of the AP.
Marking Scheme: Two equations formed 1 mark · Solving for d and a 1 mark · First three terms 1 mark.
Total: 3 Marks
Total: 3 Marks
Set up the conditions
a4 + a8 = 24
(a + 3d) + (a + 7d) = 24
2a + 10d = 24
a + 5d = 12 … (1)
(a + 3d) + (a + 7d) = 24
2a + 10d = 24
a + 5d = 12 … (1)
Second condition
a6 + a10 = 44
(a + 5d) + (a + 9d) = 44
2a + 14d = 44
a + 7d = 22 … (2)
(a + 5d) + (a + 9d) = 44
2a + 14d = 44
a + 7d = 22 … (2)
Solve (1) and (2)
(2) − (1): 2d = 10
d = 5
From (1): a + 25 = 12
a = −13
d = 5
From (1): a + 25 = 12
a = −13
First three terms
a = −13
a2 = −13 + 5 = −8
a3 = −8 + 5 = −3
a2 = −13 + 5 = −8
a3 = −8 + 5 = −3
Answer: −13, −8, −3
How many terms of the AP 9, 17, 25, … must be taken to give a sum of 636?
Marking Scheme: Correct formula and setting up the equation 1 mark · Quadratic solved 1 mark · Rejecting the invalid root and answer 1 mark.
Total: 3 Marks
Total: 3 Marks
Values and formula
a = 9, d = 8
Sn = n2[2a + (n − 1)d]
636 = n2[18 + 8(n − 1)]
Sn = n2[2a + (n − 1)d]
636 = n2[18 + 8(n − 1)]
Simplify
1272 = n(8n + 10)
1272 = 8n2 + 10n
4n2 + 5n − 636 = 0
1272 = 8n2 + 10n
4n2 + 5n − 636 = 0
Solve the quadratic
Discriminant = 52 − 4(4)(−636)
= 25 + 10176
= 10201
√10201 = 101
n = −5 ± 1018
n = 12 or n = −13.25
= 25 + 10176
= 10201
√10201 = 101
n = −5 ± 1018
n = 12 or n = −13.25
Reject the invalid root
n must be a positive whole number, so n = 12
Check: S12 = 122[18 + 88] = 6 × 106 = 636 ✓
Check: S12 = 122[18 + 88] = 6 × 106 = 636 ✓
Answer: 12 terms
The sum of three numbers in AP is 27 and their product is 648. Find the numbers.
Option 1 — Three numbers in AP
Marking Scheme: Setting the numbers as a−d, a, a+d 1 mark · Using sum and product 1 mark · Answer 1 mark.
Total: 3 Marks
Total: 3 Marks
Let the numbers be a − d, a, a + d
Sum = (a − d) + a + (a + d) = 3a
3a = 27
a = 9
3a = 27
a = 9
Use the product
(a − d)(a)(a + d) = 648
a(a2 − d2) = 648
9(81 − d2) = 648
81 − d2 = 72
d2 = 9
d = ±3
a(a2 − d2) = 648
9(81 − d2) = 648
81 − d2 = 72
d2 = 9
d = ±3
The numbers
For d = 3: 6, 9, 12
For d = −3: 12, 9, 6 (same set)
For d = −3: 12, 9, 6 (same set)
Answer: 6, 9, 12
— OR —
Option 2
Find the sum of the first 25 terms of an AP whose nth term is an = 7 − 3n.
Marking Scheme: First and last terms 1 mark · Sum formula 1 mark · Answer 1 mark.
Total: 3 Marks
Total: 3 Marks
First and 25th terms
a1 = 7 − 3 = 4
a25 = 7 − 3(25) = 7 − 75 = −68
a25 = 7 − 3(25) = 7 − 75 = −68
Sum of 25 terms
S25 = 252[a1 + a25]
= 252[4 + (−68)]
= 252 × (−64)
= −800
= 252[4 + (−68)]
= 252 × (−64)
= −800
Answer: S25 = −800
[Attempt any ONE of the two options of Q19]
The sum of the first 7 terms of an AP is 49 and the sum of the first 17 terms is 289. Find the sum of the first n terms.
Option 1 — S7 = 49, S17 = 289
Marking Scheme: Two equations 1 mark · a and d 1 mark · Sn 1 mark.
Total: 3 Marks
Total: 3 Marks
Use Sn = n/2[2a + (n − 1)d]
S7 = 72[2a + 6d] = 49
7(a + 3d) = 49
a + 3d = 7 … (1)
S17 = 172[2a + 16d] = 289
17(a + 8d) = 289
a + 8d = 17 … (2)
7(a + 3d) = 49
a + 3d = 7 … (1)
S17 = 172[2a + 16d] = 289
17(a + 8d) = 289
a + 8d = 17 … (2)
Solve
(2) − (1): 5d = 10
d = 2
From (1): a + 6 = 7
a = 1
d = 2
From (1): a + 6 = 7
a = 1
Sum of n terms
Sn = n2[2(1) + (n − 1)2]
= n2[2 + 2n − 2]
= n2 × 2n
= n2
= n2[2 + 2n − 2]
= n2 × 2n
= n2
Answer: Sn = n2
— OR —
Option 2
In an AP, the first term is 2 and the sum of the first five terms is one-fourth of the sum of the next five terms. Show that the 20th term is −112.
Marking Scheme: S5 and the next-five sum 1 mark · Equation solved for d 1 mark · a20 1 mark.
Total: 3 Marks
Total: 3 Marks
Sums in terms of d
a = 2
S5 = 52[4 + 4d] = 10 + 10d
Next five terms = S10 − S5
S10 = 102[4 + 9d] = 5(4 + 9d)
= 20 + 45d
Next five = (20 + 45d) − (10 + 10d) = 10 + 35d
S5 = 52[4 + 4d] = 10 + 10d
Next five terms = S10 − S5
S10 = 102[4 + 9d] = 5(4 + 9d)
= 20 + 45d
Next five = (20 + 45d) − (10 + 10d) = 10 + 35d
Apply the condition
10 + 10d = 14(10 + 35d)
40 + 40d = 10 + 35d
5d = −30
d = −6
40 + 40d = 10 + 35d
5d = −30
d = −6
20th term
a20 = a + 19d
= 2 + 19(−6)
= 2 − 114
= −112
= 2 + 19(−6)
= 2 − 114
= −112
Answer: a20 = −112 (shown)
[Attempt any ONE of the two options of Q20]
Section D — Case Study (2 × 4 = 8 marks)
A child builds a pyramid with tin cans. She places 1 can in the top row, 2 cans in the next row, 3 cans in the row below that, and so on. The bottom row has 20 cans.
Marking Scheme: Part (a) 1 mark · Part (b) 1.5 marks · Part (c) 1.5 marks.
Total: 4 Marks
Total: 4 Marks
(a) The AP formed
1, 2, 3, 4, …, 20
a = 1, d = 1
a = 1, d = 1
(b) Number of rows
The last term is 20 and a = 1, d = 1
an = 1 + (n − 1)(1) = n
So n = 20 rows
an = 1 + (n − 1)(1) = n
So n = 20 rows
(c) Total cans
S20 = 202 × (1 + 20)
= 10 × 21
= 210
= 10 × 21
= 210
Answer: (a) 1, 2, 3, …, 20 (b) 20 rows (c) 210 cans
Ramesh joins a savings scheme. He saves ₹100 in the first week and increases his weekly saving by ₹25 every week after that.
Option 1 — Weekly savings
Marking Scheme: Part (a) 1 mark · Part (b) 1.5 marks · Part (c) 1.5 marks.
Total: 4 Marks
Total: 4 Marks
(a) The AP formed
100, 125, 150, …
a = 100, d = 25
a = 100, d = 25
(b) Saving in the 10th week
a10 = a + 9d
= 100 + 9 × 25
= 100 + 225
= 325
= 100 + 9 × 25
= 100 + 225
= 325
(c) Total saving in 10 weeks
S10 = 102[a + a10]
= 5 × (100 + 325)
= 5 × 425
= 2125
= 5 × (100 + 325)
= 5 × 425
= 2125
Answer: (a) 100, 125, 150, … (b) ₹325 (c) ₹2125
— OR —
Option 2
A stack of logs is arranged so that the bottom row has 20 logs, the next row has 19 logs, the row above that has 18 logs, and the top row has 1 log.
Marking Scheme: Part (a) 1 mark · Part (b) 1.5 marks · Part (c) 1.5 marks.
Total: 4 Marks
Total: 4 Marks
(a) The AP formed
20, 19, 18, …, 1
a = 20, d = −1
a = 20, d = −1
(b) Number of rows
20 + (n − 1)(−1) = 1
(n − 1)(−1) = −19
n − 1 = 19
n = 20 rows
(n − 1)(−1) = −19
n − 1 = 19
n = 20 rows
(c) Total logs
S20 = 202[20 + 1]
= 10 × 21
= 210
= 10 × 21
= 210
Answer: (a) 20, 19, 18, …, 1 (b) 20 rows (c) 210 logs
[Attempt any ONE of the two options of Q22]
Self-assessment for Sections B, C and D (28 marks)
The MCQ section is scored for you. For the written questions, tick each one you got right out of the marks shown, and add them up honestly.
Written total: 0 / 28
Written and checked by Pandey Sir · Pandey Sir’s BEST Classes, Dammaiguda, Hyderabad. Every answer on this paper is worked out and checked before publishing.
This is board-pattern practice for Arithmetic Progressions, built to the current CBSE Class 10 Maths syllabus 2026-27. It is not a prediction of any examination paper.
Curriculum scope checked against cbseacademic.nic.in. Spotted a mistake? Tell us here.