Class 10 Mathematics - NCERT Mathematics
NCERT Solutions for Class 10 Maths Chapter 5: Arithmetic Progressions
Chapter 5 introduces Arithmetic Progressions (AP), where each term after the first is obtained by adding a fixed number called the common difference. Students learn to find the nth term using a_n = a + (n-1)d and the sum of first n terms using S_n = n/2[2a + (n-1)d]. This chapter is part of the Algebra unit and carries approximately 5 marks in the CBSE Class 10 board exam 2027.
Syllabus Update 2026-27
As per the CBSE 2026-27 syllabus, Chapter 5 Arithmetic Progressions has all 4 exercises fully included - Exercise 5.1 (4 questions), Exercise 5.2 (20 questions), Exercise 5.3 (20 questions), and Exercise 5.4 (5 questions). No exercises or topics have been deleted from this chapter. The chapter is part of the Algebra unit which carries 20 marks total.
Table of Contents
5.1Exercise 5.1 - Introduction to AP
Q1In which of the following situations, does the list of numbers involved make an arithmetic progression, and why? (i) The taxi fare after each km when the fare is Rs 15 for the first km and Rs 8 for each additional km. (ii) The amount of air present in a cylinder when a vacuum pump removes 1/4 of the air remaining in the cylinder at a time.
(i) Taxi fare after each km:
Fare for 1 km = Rs 15
Fare for 2 km = 15 + 8 = Rs 23
Fare for 3 km = 23 + 8 = Rs 31
Fare for 4 km = 31 + 8 = Rs 39
The list is: 15, 23, 31, 39, ...
Difference: 23-15 = 8, 31-23 = 8, 39-31 = 8
Since the common difference is constant (d = 8), this forms an AP.
(ii) Air in cylinder with vacuum pump:
Let initial air = x
After 1st pump: x - x/4 = 3x/4
After 2nd pump: 3x/4 - (3x/4)/4 = 3x/4 × 3/4 = 9x/16
After 3rd pump: 9x/16 × 3/4 = 27x/64
The list is: x, 3x/4, 9x/16, 27x/64, ...
Differences: 3x/4 - x = -x/4, 9x/16 - 3x/4 = -3x/16
Since -x/4 is not equal to -3x/16, the common difference is not constant.
This does NOT form an AP. (It forms a GP with ratio 3/4.)
Q2Write first four terms of the AP, when the first term a and the common difference d are given as follows: (i) a = 10, d = 10 (ii) a = -2, d = 0 (iii) a = 4, d = -3
(i) a = 10, d = 10:
a1 = 10, a2 = 10+10 = 20, a3 = 20+10 = 30, a4 = 30+10 = 40
AP: 10, 20, 30, 40
(ii) a = -2, d = 0:
a1 = -2, a2 = -2+0 = -2, a3 = -2+0 = -2, a4 = -2+0 = -2
AP: -2, -2, -2, -2
(iii) a = 4, d = -3:
a1 = 4, a2 = 4+(-3) = 1, a3 = 1+(-3) = -2, a4 = -2+(-3) = -5
AP: 4, 1, -2, -5
Q3For the following APs, write the first term and the common difference: (i) 3, 1, -1, -3, ... (ii) -5, -1, 3, 7, ... (iii) 1/3, 5/3, 9/3, 13/3, ...
(i) 3, 1, -1, -3, ...:
First term a = 3
Common difference d = 1 - 3 = -2
(ii) -5, -1, 3, 7, ...:
First term a = -5
Common difference d = -1 - (-5) = -1 + 5 = 4
(iii) 1/3, 5/3, 9/3, 13/3, ...:
First term a = 1/3
Common difference d = 5/3 - 1/3 = 4/3
Q4Which of the following are APs? If they form an AP, find the common difference d and write three more terms. (i) 2, 4, 8, 16, ... (ii) -1.2, -3.2, -5.2, -7.2, ... (iii) -10, -6, -2, 2, ... (iv) 1, 3, 9, 27, ...
(i) 2, 4, 8, 16, ...:
4-2 = 2, 8-4 = 4, 16-8 = 8
Differences are not equal. Not an AP. (This is a GP.)
(ii) -1.2, -3.2, -5.2, -7.2, ...:
-3.2-(-1.2) = -2, -5.2-(-3.2) = -2, -7.2-(-5.2) = -2
d = -2, constant. Yes, it is an AP.
Next three terms: -9.2, -11.2, -13.2
(iii) -10, -6, -2, 2, ...:
-6-(-10) = 4, -2-(-6) = 4, 2-(-2) = 4
d = 4, constant. Yes, it is an AP.
Next three terms: 6, 10, 14
(iv) 1, 3, 9, 27, ...:
3-1 = 2, 9-3 = 6, 27-9 = 18
Differences are not equal. Not an AP. (This is a GP.)
5.2Exercise 5.2 - nth Term of an AP
Q1Fill in the blanks in the following table, given that a is the first term, d the common difference and a_n the nth term of the AP: (i) a = 7, d = 3, n = 8, a_n = ? (ii) a = -18, d = ?, n = 10, a_n = 0
Using the formula: a_n = a + (n-1)d
(i) a = 7, d = 3, n = 8:
a_8 = 7 + (8-1)(3) = 7 + 21 = 28
(ii) a = -18, n = 10, a_n = 0:
0 = -18 + (10-1)d
0 = -18 + 9d
9d = 18
d = 2
Q3In the following APs, find the missing terms in the boxes: (i) 2, __, 26 (ii) __, 13, __, 3
(i) 2, __, 26:
Here a = 2, a3 = 26
a3 = a + 2d
26 = 2 + 2d
2d = 24
d = 12
Missing term = a + d = 2 + 12 = 14
AP: 2, 14, 26
(ii) __, 13, __, 3:
Here a2 = 13, a4 = 3
a4 = a2 + 2d
3 = 13 + 2d
2d = -10
d = -5
a1 = a2 - d = 13 - (-5) = 18
a3 = a2 + d = 13 + (-5) = 8
AP: 18, 13, 8, 3
Q4Which term of the AP: 3, 8, 13, 18, ... is 78?
Here a = 3, d = 8 - 3 = 5
Let a_n = 78
Using a_n = a + (n-1)d:
78 = 3 + (n-1)(5)
78 = 3 + 5n - 5
78 = 5n - 2
5n = 80
n = 16
78 is the 16th term of the AP.
Q10The 17th term of an AP exceeds its 10th term by 7. Find the common difference.
Let first term = a, common difference = d
a_17 = a + 16d
a_10 = a + 9d
Given: a_17 - a_10 = 7
(a + 16d) - (a + 9d) = 7
7d = 7
d = 1
The common difference is 1.
Q17Find the 20th term from the last term of the AP: 3, 8, 13, ..., 253.
The AP from the last term is: 253, 248, 243, ...
Here a' = 253, d' = -5
The 20th term from the last = a' + (20-1)d'
= 253 + 19(-5)
= 253 - 95
= 158
Alternatively: Total terms = (253-3)/5 + 1 = 250/5 + 1 = 51
20th from last = (51-20+1)th from beginning = 32nd term
a_32 = 3 + 31(5) = 3 + 155 = 158 ✓
5.3Exercise 5.3 - Sum of First n Terms of an AP
Q1Find the sum of the following APs: (i) 2, 7, 12, ..., to 10 terms (ii) -37, -33, -29, ..., to 12 terms
Using S_n = n/2[2a + (n-1)d]
(i) a = 2, d = 5, n = 10:
S_10 = 10/2[2(2) + (10-1)(5)]
= 5[4 + 45]
= 5 × 49
= 245
(ii) a = -37, d = -33-(-37) = 4, n = 12:
S_12 = 12/2[2(-37) + (12-1)(4)]
= 6[-74 + 44]
= 6 × (-30)
= -180
Q3In an AP: (i) given a = 5, d = 3, a_n = 50, find n and S_n. (ii) given a = 7, a_13 = 35, find d and S_13.
(i) a = 5, d = 3, a_n = 50:
a_n = a + (n-1)d
50 = 5 + (n-1)(3)
45 = 3(n-1)
n-1 = 15
n = 16
S_n = n/2(a + a_n) = 16/2(5 + 50) = 8 × 55 = 440
(ii) a = 7, a_13 = 35:
a_13 = a + 12d
35 = 7 + 12d
12d = 28
d = 7/3
S_13 = 13/2(a + a_13) = 13/2(7 + 35) = 13/2 × 42 = 273
Q7Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.
Given: d = 7, a_22 = 149, n = 22
First find a:
a_22 = a + 21d
149 = a + 21(7)
149 = a + 147
a = 2
S_22 = n/2(a + a_n) = 22/2(2 + 149) = 11 × 151 = 1661
Q10Show that a_1, a_2, ..., a_n, ... form an AP where a_n is defined as below: a_n = 3 + 4n. Also find the sum of the first 15 terms.
Given a_n = 3 + 4n
a_1 = 3 + 4(1) = 7
a_2 = 3 + 4(2) = 11
a_3 = 3 + 4(3) = 15
a_4 = 3 + 4(4) = 19
Difference: a_2 - a_1 = 11 - 7 = 4
a_3 - a_2 = 15 - 11 = 4
a_4 - a_3 = 19 - 15 = 4
Since the common difference is constant (d = 4), it forms an AP.
Alternatively: a_n - a_(n-1) = (3+4n) - (3+4(n-1)) = 4n - 4n + 4 = 4 (constant)
Sum of first 15 terms:
a = 7, d = 4, n = 15
S_15 = 15/2[2(7) + (15-1)(4)]
= 15/2[14 + 56]
= 15/2 × 70
= 525
Q14How many terms of the AP: 9, 17, 25, ... must be taken to give a sum of 636?
Here a = 9, d = 17 - 9 = 8, S_n = 636
Using S_n = n/2[2a + (n-1)d]:
636 = n/2[2(9) + (n-1)(8)]
636 = n/2[18 + 8n - 8]
636 = n/2[10 + 8n]
1272 = n(10 + 8n)
1272 = 10n + 8n^2
8n^2 + 10n - 1272 = 0
4n^2 + 5n - 636 = 0
Using quadratic formula:
n = [-5 + sqrt(25 + 4 × 4 × 636)] / (2 × 4)
n = [-5 + sqrt(25 + 10176)] / 8
n = [-5 + sqrt(10201)] / 8
n = (-5 + 101) / 8
n = 96/8
n = 12
(Rejecting the negative root n = -53/4)
12 terms must be taken to give a sum of 636.
Q18A ladder has rungs 25 cm apart. The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and the bottom rungs are 2.5 m apart, how many rungs are there? What is the total length of the rungs?
Distance between top and bottom = 2.5 m = 250 cm
Rungs are 25 cm apart.
Number of rungs = 250/25 + 1 = 10 + 1 = 11 rungs
Lengths form an AP: 45, ?, ?, ..., 25
a = 45 (bottom rung), last term l = 25 (top rung), n = 11
Total length = S_11 = n/2(a + l)
= 11/2(45 + 25)
= 11/2 × 70
= 11 × 35
= 385 cm
The total length of all rungs is 385 cm.
5.4Exercise 5.4 - Mixed Problems on AP
Q1Which term of the AP: 121, 117, 113, ... is its first negative term?
Here a = 121, d = 117 - 121 = -4
For the first negative term, a_n < 0:
a + (n-1)d < 0
121 + (n-1)(-4) < 0
121 - 4n + 4 < 0
125 - 4n < 0
125 < 4n
n > 31.25
Since n must be a positive integer, n = 32.
Verification: a_32 = 121 + 31(-4) = 121 - 124 = -3 (negative) ✓
a_31 = 121 + 30(-4) = 121 - 120 = 1 (positive) ✓
The 32nd term is the first negative term.
Q2The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP.
Let first term = a, common difference = d
a_3 = a + 2d, a_7 = a + 6d
Given: a_3 + a_7 = 6
(a + 2d) + (a + 6d) = 6
2a + 8d = 6
a + 4d = 3
a = 3 - 4d ... (i)
Given: a_3 × a_7 = 8
(a + 2d)(a + 6d) = 8
Substituting a = 3 - 4d:
(3 - 4d + 2d)(3 - 4d + 6d) = 8
(3 - 2d)(3 + 2d) = 8
9 - 4d^2 = 8
4d^2 = 1
d^2 = 1/4
d = 1/2 or d = -1/2
Case 1: d = 1/2, a = 3 - 4(1/2) = 1
S_16 = 16/2[2(1) + 15(1/2)] = 8[2 + 7.5] = 8 × 9.5 = 76
Case 2: d = -1/2, a = 3 - 4(-1/2) = 5
S_16 = 16/2[2(5) + 15(-1/2)] = 8[10 - 7.5] = 8 × 2.5 = 20
The sum of first 16 terms is 76 or 20.
Q3A manufacturer of TV sets produced 600 sets in the third year and 700 sets in the seventh year. Assuming that the production increases uniformly by a fixed number every year, find: (i) the production in the 1st year (ii) the production in the 10th year (iii) the total production in first 7 years.
Production forms an AP with first term a and common difference d.
Given: a_3 = 600, a_7 = 700
a + 2d = 600 ... (i)
a + 6d = 700 ... (ii)
Subtracting (i) from (ii):
4d = 100
d = 25 (production increases by 25 sets each year)
From (i): a = 600 - 2(25) = 600 - 50 = 550
(i) Production in 1st year = a = 550 sets
(ii) Production in 10th year:
a_10 = a + 9d = 550 + 9(25) = 550 + 225 = 775 sets
(iii) Total production in first 7 years:
S_7 = 7/2[2(550) + 6(25)]
= 7/2[1100 + 150]
= 7/2 × 1250
= 4375 sets
Q4A sum of Rs 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is Rs 20 less than its preceding prize, find the value of each of the prizes.
Let the first (highest) prize = a
Since each prize is Rs 20 less than the preceding one, d = -20.
There are 7 prizes, so n = 7.
S_7 = 700
7/2[2a + 6(-20)] = 700
7/2[2a - 120] = 700
7(2a - 120) = 1400
2a - 120 = 200
2a = 320
a = 160
The seven prizes are:
160, 140, 120, 100, 80, 60, 40
Verification: 160+140+120+100+80+60+40 = 700 ✓
The prizes are Rs 160, Rs 140, Rs 120, Rs 100, Rs 80, Rs 60, and Rs 40.
Q5200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on. In how many rows are the 200 logs placed and how many logs are in the top row?
Number of logs form an AP: 20, 19, 18, ...
Here a = 20, d = -1, S_n = 200
Using S_n = n/2[2a + (n-1)d]:
200 = n/2[2(20) + (n-1)(-1)]
200 = n/2[40 - n + 1]
400 = n(41 - n)
400 = 41n - n^2
n^2 - 41n + 400 = 0
Using quadratic formula:
n = [41 + sqrt(1681 - 1600)] / 2
n = [41 + sqrt(81)] / 2
n = (41 + 9) / 2 or n = (41 - 9) / 2
n = 25 or n = 16
Since a_25 = 20 + 24(-1) = -4 (negative, not possible),
n = 16 rows
Logs in top row = a_16 = 20 + 15(-1) = 20 - 15 = 5 logs
Verification: S_16 = 16/2(20 + 5) = 8 × 25 = 200 ✓
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Ask AI TutorPrevious Year Board Exam Questions
If the nth term of an AP is 7 - 4n, find its common difference.
Given a_n = 7 - 4n. The first term a_1 = 7 - 4(1) = 3. The second term a_2 = 7 - 4(2) = -1. Common difference d = a_2 - a_1 = -1 - 3 = -4. Alternatively, since a_n = 7 - 4n is linear in n, the coefficient of n gives the common difference directly. So d = -4.
How many two-digit numbers are divisible by 3?
Two-digit numbers divisible by 3 form an AP: 12, 15, 18, ..., 99. Here first term a = 12, common difference d = 3, last term l = 99. Using a_n = a + (n-1)d, we get 99 = 12 + (n-1)(3), so 87 = 3(n-1), giving n-1 = 29, hence n = 30. There are 30 two-digit numbers divisible by 3.
Find the sum of first 20 terms of the AP: 1, 4, 7, 10, ...
Here a = 1, d = 4 - 1 = 3, n = 20. Using S_n = n/2[2a + (n-1)d], we get S_20 = 20/2[2(1) + 19(3)] = 10[2 + 57] = 10 x 59 = 590. The sum of the first 20 terms is 590.
Which term of the AP: 3, 15, 27, 39, ... will be 120 more than its 21st term?
Here a = 3, d = 15 - 3 = 12. The 21st term a_21 = 3 + 20(12) = 3 + 240 = 243. We need a_n = 243 + 120 = 363. Using a_n = a + (n-1)d: 363 = 3 + (n-1)(12), so 360 = 12(n-1), giving n-1 = 30, hence n = 31. The 31st term will be 120 more than the 21st term.
Find the number of terms of the AP: 54, 51, 48, ... so that their sum is 513.
Here a = 54, d = 51 - 54 = -3, S_n = 513. Using S_n = n/2[2a + (n-1)d]: 513 = n/2[108 + (n-1)(-3)] = n/2[108 - 3n + 3] = n/2[111 - 3n]. So 1026 = n(111 - 3n) = 111n - 3n^2. Rearranging: 3n^2 - 111n + 1026 = 0, or n^2 - 37n + 342 = 0. Factoring: (n - 18)(n - 19) = 0. So n = 18 or n = 19. Both answers are valid because a_19 = 54 + 18(-3) = 0, so adding the 19th term (which is 0) does not change the sum.
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.
Given a = 5, l = 45, S_n = 400. Using S_n = n/2(a + l): 400 = n/2(5 + 45) = n/2(50) = 25n. So n = 400/25 = 16. Now using l = a + (n-1)d: 45 = 5 + 15d, so 15d = 40, giving d = 40/15 = 8/3. The number of terms is 16 and the common difference is 8/3.
A contract on construction job specifies a penalty for delay of completion beyond a certain date. Rs 200 for the first day, Rs 250 for the second day, Rs 300 for the third day, etc. If the contractor pays Rs 27750 as penalty, find the number of days for which the construction work was delayed.
Penalties form an AP: 200, 250, 300, ... with a = 200, d = 50, S_n = 27750. Using S_n = n/2[2a + (n-1)d]: 27750 = n/2[400 + 50(n-1)] = n/2[400 + 50n - 50] = n/2[350 + 50n]. So 55500 = n(350 + 50n) = 350n + 50n^2. Dividing by 50: n^2 + 7n - 1110 = 0. Factoring: (n + 37)(n - 30) = 0. Since n must be positive, n = 30. The construction was delayed by 30 days.
Find the value of k for which the numbers (3k - 2), (4k - 6), and (k + 2) are three consecutive terms of an AP.
If three numbers are consecutive terms of an AP, the middle term is the average of the other two. So 2(4k - 6) = (3k - 2) + (k + 2). Simplifying: 8k - 12 = 3k - 2 + k + 2 = 4k. So 8k - 12 = 4k, giving 4k = 12, hence k = 3. Verification: terms are 7, 6, 5 with d = -1. This is indeed an AP.
If the sum of the first 14 terms of an AP is 1050 and its first term is 10, find the 20th term.
Given S_14 = 1050, a = 10. Using S_n = n/2[2a + (n-1)d]: 1050 = 14/2[2(10) + 13d] = 7[20 + 13d]. So 150 = 20 + 13d, giving 13d = 130, hence d = 10. The 20th term: a_20 = a + 19d = 10 + 19(10) = 10 + 190 = 200. The 20th term is 200.
Jaspal Singh repays his total loan of Rs 118000 by paying every month starting with the first instalment of Rs 1000. If he increases the instalment by Rs 100 every month, what amount will be paid as the last instalment of the loan?
Monthly instalments form an AP: 1000, 1100, 1200, ... with a = 1000, d = 100. Total loan S_n = 118000. Using S_n = n/2[2a + (n-1)d]: 118000 = n/2[2000 + 100(n-1)] = n/2[2000 + 100n - 100] = n/2[1900 + 100n]. So 236000 = n(1900 + 100n) = 1900n + 100n^2. Dividing by 100: n^2 + 19n - 2360 = 0. Factoring: (n + 59)(n - 40) = 0. So n = 40 months. Last instalment = a_40 = 1000 + 39(100) = 1000 + 3900 = Rs 4900.
Find the sum of the AP: -5 + (-2) + 1 + 4 + ... + 73.
Here a = -5, d = -2 - (-5) = 3, l = 73. First find n using a_n = a + (n-1)d: 73 = -5 + (n-1)(3), so 78 = 3(n-1), giving n-1 = 26, hence n = 27. Now S_n = n/2(a + l) = 27/2(-5 + 73) = 27/2(68) = 27 x 34 = 918. The sum of the AP is 918.
The first term of an AP is -5 and the last term is 45. If the sum of the terms of the AP is 120, then find the number of terms and the common difference.
Given a = -5, l = 45, S_n = 120. Using S_n = n/2(a + l): 120 = n/2(-5 + 45) = n/2(40) = 20n. So n = 120/20 = 6. Now using l = a + (n-1)d: 45 = -5 + 5d, so 5d = 50, giving d = 10. The AP has 6 terms and the common difference is 10.
The 4th term of an AP is zero. Prove that the 25th term of the AP is three times its 11th term.
Given a_4 = 0, so a + 3d = 0, which gives a = -3d. Now a_25 = a + 24d = -3d + 24d = 21d. And a_11 = a + 10d = -3d + 10d = 7d. We need to show a_25 = 3 x a_11: 3 x a_11 = 3(7d) = 21d = a_25. Hence proved that the 25th term is three times the 11th term.
Find the sum of first 24 terms of an AP whose nth term is given by a_n = 3 + 2n.
Given a_n = 3 + 2n. First term a_1 = 3 + 2(1) = 5. Last term (24th) a_24 = 3 + 2(24) = 51. Using S_n = n/2(a + l): S_24 = 24/2(5 + 51) = 12(56) = 672. The sum of the first 24 terms is 672.
If the 2nd term of an AP is 13 and the 5th term is 25, what is its 7th term?
Given a_2 = 13 and a_5 = 25. So a + d = 13 ... (i) and a + 4d = 25 ... (ii). Subtracting (i) from (ii): 3d = 12, d = 4. From (i): a = 13 - 4 = 9. The 7th term a_7 = a + 6d = 9 + 24 = 33. The 7th term is 33.
Find the sum of first 100 even natural numbers which are divisible by 5.
Even natural numbers divisible by 5 are: 10, 20, 30, 40, ... This is an AP with a = 10, d = 10. The 100th term a_100 = 10 + 99(10) = 1000. Sum S_100 = 100/2(10 + 1000) = 50 x 1010 = 50500. The sum of the first 100 such numbers is 50500.
Want more PYQ-style practice? AI generates unlimited similar questions from this chapter.
Practice PYQ PatternsAssertion-Reason Questions
(A) Both true, Reason explains Assertion · (B) Both true, Reason doesn't explain · (C) Assertion true, Reason false · (D) Assertion false, Reason true
AR1
Assertion (A): The common difference of the AP: -5, -1, 3, 7, ... is 4.
Reason (R): The common difference of an AP is given by d = a_2 - a_1.
Answer: (A)
Both are true. d = a_2 - a_1 = -1 - (-5) = -1 + 5 = 4. The reason correctly explains the method to find the common difference, and it correctly gives d = 4.
AR2
Assertion (A): The nth term of the AP: 3, 8, 13, 18, ... is 5n - 2.
Reason (R): The nth term of an AP is given by a_n = a + (n-1)d.
Answer: (A)
Both are true and the reason explains the assertion. Here a = 3, d = 5. So a_n = 3 + (n-1)(5) = 3 + 5n - 5 = 5n - 2. The reason provides the formula used to derive the nth term.
AR3
Assertion (A): The 11th term of the AP: -3, -1/2, 2, ... is 22.
Reason (R): The nth term of an AP with first term a and common difference d is a_n = a + (n-1)d.
Answer: (A)
Both are true. Here a = -3, d = -1/2 - (-3) = 5/2. a_11 = -3 + 10(5/2) = -3 + 25 = 22. The reason correctly explains the formula used.
AR4
Assertion (A): The sum of the first 1000 positive integers is 500500.
Reason (R): The sum of first n natural numbers is S_n = n(n+1)/2.
Answer: (A)
Both are true. S_1000 = 1000(1001)/2 = 500500. The reason gives the correct formula (which is derived from the AP sum formula with a = 1, d = 1), and it correctly explains the assertion.
AR5
Assertion (A): The sum of the first 10 terms of the AP: 2, 7, 12, ... is 250.
Reason (R): S_n = n/2(a + l) where l is the last term.
Answer: (D)
Reason is TRUE as a general formula. But Assertion is FALSE: a = 2, d = 5, n = 10. S_10 = 10/2[2(2) + 9(5)] = 5[4 + 45] = 5 x 49 = 245, not 250.
AR6
Assertion (A): If the sum of the first n terms of an AP is 3n^2 + 5n, then the common difference is 6.
Reason (R): The common difference can be found using d = S_2 - 2S_1.
Answer: (A)
Both are true. S_1 = 3(1) + 5 = 8 = a_1. S_2 = 3(4) + 10 = 22. a_2 = S_2 - S_1 = 22 - 8 = 14. d = a_2 - a_1 = 14 - 8 = 6. The reason gives a valid method: d = S_2 - 2S_1 = 22 - 16 = 6.
AR7
Assertion (A): The 10th term of the AP: 5, 8, 11, 14, ... is __(32)__.
Reason (R): The sum of the first n terms of an AP is S_n = n/2[2a + (n-1)d].
Answer: (B)
Assertion is TRUE: a = 5, d = 3, a_10 = 5 + 9(3) = 32. Reason is also TRUE as a formula. However, the reason is about sum of terms, not about finding the nth term. Since the reason does not explain the assertion, the answer is B.
Case Study Questions
In a school stadium, seats are arranged in rows. The first row has 25 seats, the second row has 27 seats, the third row has 29 seats, and so on. Each row has 2 more seats than the previous row. The stadium has a total of 30 rows.
(i) How many seats are there in the 15th row? (1 mark)
This is an AP with a = 25, d = 2.
a_15 = a + 14d = 25 + 14(2) = 25 + 28 = 53 seats in the 15th row.
(ii) How many seats are there in the 30th (last) row? (1 mark)
a_30 = a + 29d = 25 + 29(2) = 25 + 58 = 83 seats in the 30th row.
(iii)(a) What is the total number of seats in the stadium? (2 marks)
Total seats = S_30 = 30/2(a + a_30) = 15(25 + 83) = 15 x 108 = 1620 seats in total.
(iii)(b) How many rows are needed to seat at least 500 people? (2 marks)
We need S_n >= 500.
S_n = n/2[2(25) + (n-1)(2)] = n/2[50 + 2n - 2] = n/2[48 + 2n] = n(24 + n) = n^2 + 24n.
Solving n^2 + 24n = 500: n^2 + 24n - 500 = 0.
Using the quadratic formula: n = [-24 + sqrt(576 + 2000)] / 2 = [-24 + sqrt(2576)] / 2 = [-24 + 50.75] / 2 = 13.38.
Since n must be a whole number, n = 14 rows are needed to seat at least 500 people.
Verification: S_14 = 14(24 + 14) = 14(38) = 532 >= 500 and S_13 = 13(37) = 481 < 500.
Priya decides to save money for buying a bicycle that costs Rs 10000. She saves Rs 100 in the first month, Rs 150 in the second month, Rs 200 in the third month, and so on, increasing her savings by Rs 50 each month.
(i) How much money does Priya save in the 10th month? (1 mark)
Savings form an AP with a = 100, d = 50.
a_10 = 100 + 9(50) = 100 + 450 = Rs 550 in the 10th month.
(ii) What is the total savings after 12 months? (1 mark)
S_12 = 12/2[2(100) + 11(50)] = 6[200 + 550] = 6 x 750 = Rs 4500 total after 12 months.
(iii)(a) After how many months will Priya have saved enough to buy the bicycle? (2 marks)
We need S_n >= 10000.
S_n = n/2[200 + 50(n-1)] = n/2[200 + 50n - 50] = n/2[150 + 50n] = n(75 + 25n) = 25n^2 + 75n.
Solving 25n^2 + 75n = 10000: n^2 + 3n - 400 = 0.
Using quadratic formula: n = [-3 + sqrt(9 + 1600)] / 2 = [-3 + sqrt(1609)] / 2 = [-3 + 40.11] / 2 = 18.56.
Since n must be a whole number, n = 19 months.
Verification: S_19 = 25(361) + 75(19) = 9025 + 1425 = 10450 >= 10000 and S_18 = 25(324) + 75(18) = 8100 + 1350 = 9450 < 10000.
(iii)(b) What is the total amount saved in the first 8 months? (2 marks)
S_8 = 8/2[2(100) + 7(50)] = 4[200 + 350] = 4 x 550 = Rs 2200 saved in the first 8 months.
MCQ Practice
MCQ1. The common difference of the AP: 1/p, (1-p)/p, (1-2p)/p, ... is:
(A) 1
(B) 1/p
(C) -1
(D) -1/p
Answer: (C) - d = (1-p)/p - 1/p = (1-p-1)/p = -p/p = -1.
MCQ2. The 10th term of the AP: 2, 7, 12, ... is:
(A) 45
(B) 47
(C) 49
(D) 50
Answer: (B) - a = 2, d = 5. a_10 = 2 + 9(5) = 2 + 45 = 47.
MCQ3. The first term of an AP is p and the common difference is q, then its 10th term is:
(A) q + 9p
(B) p - 9q
(C) p + 9q
(D) 2p + 9q
Answer: (C) - a_10 = a + (10-1)d = p + 9q.
MCQ4. If the nth term of an AP is (2n + 1), the sum of first n terms of the AP is:
(A) n(n + 2)
(B) n(n + 1)
(C) n(n - 1)
(D) n(n + 3)
Answer: (A) - a_1 = 3, a_n = 2n+1. S_n = n/2(a_1 + a_n) = n/2(3 + 2n + 1) = n/2(2n + 4) = n(n + 2).
MCQ5. The sum of first 16 terms of the AP: 10, 6, 2, ... is:
(A) -320
(B) 320
(C) -352
(D) -400
Answer: (A) - a = 10, d = -4, n = 16. S_16 = 16/2[2(10) + 15(-4)] = 8[20 - 60] = 8(-40) = -320.
MCQ6. If the 2nd term of an AP is 13 and the 5th term is 25, then the common difference is:
(A) 3
(B) 4
(C) 5
(D) 6
Answer: (B) - a_2 = a + d = 13, a_5 = a + 4d = 25. Subtracting: 3d = 12, d = 4.
MCQ7. Which term of the AP: 21, 18, 15, ... is -81?
(A) 33rd
(B) 34th
(C) 35th
(D) 36th
Answer: (C) - a = 21, d = -3. a_n = 21 + (n-1)(-3) = -81. So 21 - 3n + 3 = -81, 24 - 3n = -81, 3n = 105, n = 35.
MCQ8. If the first term of an AP is -5 and the common difference is 2, then the sum of the first 6 terms is:
(A) 0
(B) 5
(C) 6
(D) 15
Answer: (A) - a = -5, d = 2, n = 6. S_6 = 6/2[2(-5) + 5(2)] = 3[-10 + 10] = 3(0) = 0.
MCQ9. The number of two-digit numbers which are divisible by 6 is:
(A) 14
(B) 15
(C) 16
(D) 17
Answer: (B) - Two-digit numbers divisible by 6: 12, 18, 24, ..., 96. a = 12, d = 6, l = 96. Using 96 = 12 + (n-1)6: 84 = 6(n-1), n-1 = 14, n = 15.
MCQ10. The sum of first 200 natural numbers is:
(A) 20000
(B) 20100
(C) 20200
(D) 19900
Answer: (B) - S_n = n(n+1)/2 = 200(201)/2 = 100 x 201 = 20100.
MCQ11. If a, b, c are in AP, then which of the following is true?
(A) b = a + c
(B) 2b = a + c
(C) b^2 = a + c
(D) 2a = b + c
Answer: (B) - If a, b, c are in AP, then b - a = c - b. So 2b = a + c. The middle term is the arithmetic mean of the other two.
MCQ12. The 21st term of the AP whose first two terms are -3 and 4 is:
(A) 17
(B) 137
(C) 143
(D) -143
Answer: (B) - a = -3, d = 4 - (-3) = 7. a_21 = -3 + 20(7) = -3 + 140 = 137.
Try Quiz mode - 10 MCQs per topic with instant feedback and scoring.
Start Quiz ModeImportant Questions - Mark-wise
1 Mark Questions
What is the common difference of the AP: 1/3, 5/3, 9/3, 13/3, ...?
d = 5/3 - 1/3 = 4/3. The common difference is 4/3.
If the nth term of an AP is 7 - 4n, what is the common difference?
a_1 = 7 - 4 = 3, a_2 = 7 - 8 = -1. d = -1 - 3 = -4. Alternatively, in a_n = 7 - 4n, the coefficient of n gives d = -4.
Write the first four terms of the AP where a = -1 and d = 1/2.
-1, -1/2, 0, 1/2. Each term is obtained by adding 1/2 to the previous term.
How many two-digit numbers are divisible by 7?
AP: 14, 21, 28, ..., 98. Using 98 = 14 + (n-1)(7): 84 = 7(n-1), n = 13. There are 13 two-digit numbers divisible by 7.
If 4, x, 16 are in AP, find the value of x.
If three numbers are in AP, the middle term equals the average of the other two. So x = (4 + 16)/2 = 10.
2-3 Mark Questions
Find the 20th term from the last term of the AP: 3, 8, 13, ..., 253.
Total terms: 253 = 3 + (n-1)(5), so n = 51. The 20th term from last = (51-20+1) = 32nd from start. a_32 = 3 + 31(5) = 158.
Find the sum of the first 15 multiples of 8.
AP: 8, 16, 24, ..., 120. S_15 = 15/2(8 + 120) = 15/2(128) = 15 x 64 = 960.
The 17th term of an AP exceeds its 10th term by 7. Find the common difference.
a_17 - a_10 = (a + 16d) - (a + 9d) = 7d = 7. So d = 1.
Find the value of k if (2k + 1), (k^2 + k + 1), and (3k^2 - 3k + 3) are in AP.
For AP: 2(k^2+k+1) = (2k+1) + (3k^2-3k+3). So 2k^2+2k+2 = 3k^2-k+4. Rearranging: k^2-3k+2 = 0, (k-1)(k-2) = 0. k = 1 or k = 2.
3-4 Mark Questions
How many terms of the AP: 9, 17, 25, ... must be taken to give a sum of 636?
a = 9, d = 8, S_n = 636. Using S_n = n/2[2a+(n-1)d]: 636 = n/2[18+8n-8] = n/2[10+8n]. So 1272 = 10n+8n^2, or 4n^2+5n-636 = 0. n = (-5+101)/8 = 12.
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.
S_n = n/2(a+l): 400 = n/2(50), n = 16. Using l = a+(n-1)d: 45 = 5+15d, d = 40/15 = 8/3.
200 logs are stacked such that 20 logs are in the bottom row, 19 in the next, and so on. In how many rows are the 200 logs placed and how many logs are in the top row?
AP: 20, 19, 18, ... with S_n = 200. n/2[40-n+1] = 200, so n(41-n) = 400. n^2-41n+400 = 0, (n-16)(n-25) = 0. n = 16 (since a_25 would be negative). Logs in top row = 20-15 = 5.
Key Concepts
Important Formulas
Study Plan - Arithmetic Progressions
Introduction to AP and Common Difference
Read NCERT theory on what an AP is. Understand common difference with examples. Solve all 4 questions of Exercise 5.1. Identify which sequences are AP and which are not.
nth Term Formula
Learn the formula a_n = a + (n-1)d. Solve Exercise 5.2 Q1-Q10. Focus on finding specific terms, missing terms, and which term equals a given value.
Advanced nth Term Problems
Solve Exercise 5.2 Q11-Q20. Practice word problems on nth term, finding terms from the end, and problems where two APs are compared.
Sum of First n Terms
Learn S_n = n/2[2a + (n-1)d] and S_n = n/2(a + l). Solve Exercise 5.3 Q1-Q10. Practice direct sum calculations and finding n given the sum.
Word Problems on Sum and Mixed Problems
Solve Exercise 5.3 Q11-Q20 and all 5 questions of Exercise 5.4. Focus on real-life word problems, first negative term, and combined nth term-sum problems.
Revision, PYQs, and Exam Practice
Solve previous year questions (2019-2026). Practice MCQs, assertion-reason, and case study questions. Revise all formulas. Time yourself - aim for 3-mark questions in 5 minutes.
Common Mistakes to Avoid
- ✗Confusing the nth term formula a_n = a + (n-1)d with a_n = a + nd - always remember to subtract 1 from n before multiplying by d
- ✗Using S_n = n/2(a + l) when the last term is unknown - use S_n = n/2[2a + (n-1)d] instead when you do not know the last term
- ✗Forgetting to check both roots when solving quadratic equations from S_n problems - always reject negative or non-integer values of n
- ✗Writing d = a_1 - a_2 instead of d = a_2 - a_1 - the common difference is always the next term minus the current term, not the other way round
- ✗Not verifying whether a sequence is actually an AP before applying formulas - check that the difference between consecutive terms is constant for at least three pairs
- ✗In word problems, confusing whether to find a_n (specific term) or S_n (sum of terms) - read the question carefully to determine if it asks for a single value or a total
Tips for Board Exam
- ✓Memorise both sum formulas: S_n = n/2[2a + (n-1)d] for when you know a and d, and S_n = n/2(a + l) for when you know the first and last terms - choosing the right one saves time
- ✓For 'how many terms' questions that lead to quadratic equations, always check both roots and reject the one that gives a negative number of terms or a non-integer
- ✓When three terms a, b, c are in AP, use the shortcut 2b = a + c - this is faster than setting up d = b - a = c - b and solving
- ✓In word problems, always identify a, d, and n clearly before writing the formula - label them in your rough work to avoid substitution errors
- ✓For questions like 'how many 2-digit numbers are divisible by k', the AP is always k, 2k, ... up to the largest 2-digit multiple, and you can quickly find n using (last - first)/d + 1
Practice Arithmetic Progressions with AI Tutor
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Start Arithmetic Progressions - Choose Your ModeAbout Arithmetic Progressions - Class 10 Mathematics
Chapter 5 Arithmetic Progressions is one of the most important chapters in CBSE Class 10 Mathematics. It belongs to the Algebra unit, which carries 20 marks in total. AP alone typically contributes around 5 marks in the board exam, often appearing as one MCQ, one short-answer question, and sometimes as part of a case study. The chapter builds on the concept of sequences and patterns that students have encountered in earlier classes, formalising them into a powerful algebraic framework.
What Is an Arithmetic Progression?
An Arithmetic Progression (AP) is a sequence of numbers in which each term after the first is obtained by adding a fixed number to the previous term. This fixed number is called the common difference and is denoted by d. For example, the sequence 2, 5, 8, 11, 14, ... is an AP with first term a = 2 and common difference d = 3. APs appear everywhere in daily life - the fare structure of a taxi that charges a base fare plus a fixed amount per kilometre, the number of seats in successive rows of a stadium, monthly savings that increase by a fixed amount, and salary increments that follow a fixed annual raise. Not every sequence is an AP. The sequence 1, 4, 9, 16, ... (perfect squares) is not an AP because the differences between consecutive terms are 3, 5, 7, ... which are not constant. Similarly, a geometric sequence like 2, 6, 18, 54, ... has a constant ratio but not a constant difference.
Common Difference
The common difference d of an AP is found by subtracting any term from the term that follows it: d = a_(n+1) - a_n. This value remains the same throughout the AP. The common difference can be positive (increasing AP like 3, 7, 11, 15, ...), negative (decreasing AP like 20, 17, 14, 11, ...), or zero (constant AP like 5, 5, 5, 5, ...). To check whether a given sequence forms an AP, compute the differences between consecutive terms. If all differences are equal, it is an AP. For example, for the sequence -2, 2, 6, 10: differences are 4, 4, 4, so it is an AP with d = 4. For the sequence 1, 1, 2, 3, 5: differences are 0, 1, 1, 2, which are not constant, so it is not an AP.
The nth Term Formula
The nth term (or general term) of an AP is given by a_n = a + (n-1)d, where a is the first term, d is the common difference, and n is the position of the term. This formula allows you to find any term of the AP without listing all previous terms. For example, to find the 50th term of the AP 3, 7, 11, ...: a_50 = 3 + 49(4) = 3 + 196 = 199. The formula also helps in reverse. If you know a particular term, you can find the position. For instance, which term of the AP 5, 9, 13, ... is 101? Setting 101 = 5 + (n-1)(4) gives n = 25. Another common application is finding a term from the end of a finite AP. If the last term is l, the nth term from the end is l - (n-1)d. The nth term formula is also used to check whether a given number belongs to an AP. If solving for n gives a positive integer, the number is a term of the AP; otherwise, it is not.
Sum of First n Terms
The sum of the first n terms of an AP is denoted by S_n and is calculated using two equivalent formulas. When the first term a and common difference d are known: S_n = n/2[2a + (n-1)d]. When the last term l is known: S_n = n/2(a + l). The second formula is faster when the last term is directly available. For example, the sum of the first 20 terms of the AP 1, 4, 7, 10, ...: S_20 = 20/2[2(1) + 19(3)] = 10(2 + 57) = 10(59) = 590. A useful relationship connects S_n and a_n: a_n = S_n - S_(n-1) for n >= 2, and a_1 = S_1. This means if you are given S_n as a formula in terms of n, you can extract the nth term. For example, if S_n = 3n^2 + 5n, then a_n = S_n - S_(n-1) = 3n^2 + 5n - 3(n-1)^2 - 5(n-1) = 6n + 2. Problems involving S_n often lead to quadratic equations. Always solve for n using the quadratic formula or factoring and reject any negative or non-integer solution since n must be a positive integer.
Word Problems on AP
Word problems form a significant portion of CBSE board questions on AP. Common types include salary or savings problems where amounts increase by a fixed value each month or year, production problems where output increases uniformly, seating arrangements where each row has a fixed number of extra seats, penalty problems where fines increase daily by a fixed amount, and stacking problems where layers decrease uniformly. The key to solving word problems is identifying the first term a, the common difference d, and whether the question asks for a_n (a specific term) or S_n (total sum). For example, if a contractor faces a penalty of Rs 200 for the first day of delay and Rs 50 more for each subsequent day, the penalties form an AP with a = 200, d = 50. If the total penalty is Rs 27750, using the sum formula gives the number of days of delay.
How Tested in Board Exams
In CBSE board exams, AP questions typically appear in several formats. MCQs (1 mark each) test basic concepts like finding d, the nth term, or the sum. Short-answer questions (2-3 marks) involve finding specific terms, the number of terms, or straightforward sums. Long-answer questions (3-4 marks) feature word problems or multi-step calculations. Since 2024, case study questions (4-5 marks) on AP have appeared, usually set in real-life contexts like stadium seating, savings plans, or production scenarios. Assertion-reason questions test conceptual understanding of AP properties. The most frequently asked question types are finding the number of two-digit numbers divisible by a given number, finding the sum of first n terms, word problems on savings or production, and finding k when three expressions involving k are in AP.
Tips for Full Marks
To score full marks on AP questions, always write the formula before substituting values, as CBSE awards marks for showing the method. Verify your answer by back-substitution when possible. For sum problems that yield two values of n, check both and explain why one is rejected. In word problems, clearly define what a, d, and n represent. Use S_n = n/2(a + l) whenever the last term is known, as it reduces calculation. For MCQs, the shortcut d = coefficient of n in a linear a_n expression saves time. Always double-check the sign of d in problems involving decreasing sequences.
Connection to Other Chapters
Arithmetic Progressions connects to several other chapters in Class 10 Mathematics. The quadratic equations from Chapter 4 are needed when solving S_n problems that lead to quadratic equations. Coordinate geometry (Chapter 7) uses AP when dealing with equally spaced collinear points. In statistics (Chapter 13), the assumed mean method for finding the mean of grouped data uses concepts related to equally spaced class marks that form an AP. Understanding AP also lays the foundation for geometric progressions and series studied in Class 11. The algebraic skills of forming and solving equations, which run through Chapters 3, 4, and 5, are all interconnected.
Frequently Asked Questions
How many exercises are in NCERT Class 10 Maths Chapter 5 Arithmetic Progressions for CBSE 2026-27?
Chapter 5 Arithmetic Progressions has 4 exercises in the CBSE 2026-27 syllabus: Exercise 5.1 (4 questions on identifying AP and finding common difference), Exercise 5.2 (20 questions on the nth term formula), Exercise 5.3 (20 questions on sum of first n terms), and Exercise 5.4 (5 mixed problems). All exercises are fully included with no deletions.
What is the formula for the nth term of an AP?
The nth term of an AP is given by a_n = a + (n-1)d, where a is the first term, d is the common difference, and n is the position of the term. For example, the 10th term of the AP 2, 5, 8, ... is a_10 = 2 + 9(3) = 29.
How do you find the sum of first n terms of an AP?
Use S_n = n/2[2a + (n-1)d] when you know the first term and common difference. Use S_n = n/2(a + l) when you know both the first and last terms. For example, the sum of first 10 terms of the AP 3, 7, 11, ... is S_10 = 10/2[2(3) + 9(4)] = 5(42) = 210.
How many marks does Chapter 5 Arithmetic Progressions carry in the CBSE Class 10 board exam?
Arithmetic Progressions is part of the Algebra unit which carries 20 marks total. Chapter 5 typically contributes about 5 marks, usually as 1 MCQ (1 mark) and 1 short-answer question (2-3 marks), plus it may appear in case study questions.
What are the most important questions from Chapter 5 for the CBSE board exam?
The most frequently asked topics are: finding the number of two-digit or three-digit numbers divisible by a given number, finding the sum of first n terms, word problems on savings or production that use AP, finding k when three expressions are in AP, and proving relationships between specific terms of an AP.
How do you check if a sequence is an AP?
Calculate the differences between consecutive terms. If the difference is the same for every pair of consecutive terms, the sequence is an AP. For example, 3, 7, 11, 15 has differences 4, 4, 4 (constant), so it is an AP. But 1, 4, 9, 16 has differences 3, 5, 7 (not constant), so it is not an AP.
What is the difference between the nth term and the sum of n terms in an AP?
The nth term (a_n) gives the value of a single specific term at position n. The sum of n terms (S_n) gives the total of all terms from the 1st to the nth. They are related by a_n = S_n - S_(n-1) for n >= 2. For example, in the AP 2, 5, 8, the 3rd term a_3 = 8, but S_3 = 2 + 5 + 8 = 15.
Can the common difference of an AP be negative or zero?
Yes. If d is positive, the AP is increasing (e.g., 3, 7, 11, 15). If d is negative, the AP is decreasing (e.g., 20, 17, 14, 11). If d is zero, all terms are equal (e.g., 5, 5, 5, 5). All three are valid APs.