Arithmetic Progressions (A.P.) help students understand patterns in numbers and develop strong problem-solving skills. Chapter 5: Arithmetic Progressions focuses on the concept of an A.P., finding the nth term, calculating the sum of n terms, and solving real-life problems based on sequences and series.
These NCERT Solutions for Class 10 Maths Chapter 5 include important CBSE board questions asked between 2019 and 2025. All solutions are explained step by step in simple language to help students understand concepts clearly and score well in board exams.
NCERT Solutions for Class 10 Maths Chapter 5 – Arithmetic Progressions
Q.
How many three-digit numbers are divisible by 7?
Q.
The sum of four consecutive numbers in AP is 32 and the ratio of the product of the first and last terms to the product of two middle terms is 7 : 15. Find the numbers.
[CBSE - 2020]
Q.
The sum of the first 'm' terms of an Arithmetic Progression (AP) is given by the expression
2m2+3m. Determine the second term of this AP.
Q.
250 logs are stacked in the following manner :
22 logs in the bottom row, 21 in the next row, 20 in the row next to it and so on (as shown by an example). In how many rows, are the 250 logs placed and how many logs are there in the top row?
[CBSE - 2023]
Q.
The ratio of the
11th term to
17th term of an A.P. is 3 : 4. Find the ratio of
5th term to
21st term of the same A.P. Also, find the ratio of the sum of first 5 terms to that of first 21 terms.
[CBSE - 2023]
Q.
In an A.P. of 40 terms, the sum of first 9 terms is 153 and the sum of last 6 terms is 687. Determine the first term and common difference of A.P. Also, find the sum of all the terms of the A.P.
[CBSE - 2024]
Q.
The sum of first and eighth terms of an A.P. is 32 and their product is 60 . Find the first term and common difference of the A.P. Hence, also find the sum of its first 20 terms.
[CBSE - 2024]
Q.
If the sum of first four terms of an AP is 40 and that of first 14 terms is 280. Find the sum of its first n terms.
[CBSE - 2019]
Q.
Solve :
1+4+7+10+…+x=287
[CBSE - 2020]
Q.
A school is organizing a charity run to raise funds for a local hospital. The run is planned as a series of rounds around a track, with each round being 300 metres. To make the event more challenging and engaging, the organizers decide to increase the distance of each subsequent round by 50 metres. For example, the second round will be 350 metres, the third round will be 400 metres and so on. The total number of rounds planned is 10.
Based on the information given above, answer the following questions :
(i) Write the fourth, fifth and sixth term of the Arithmetic Progression so formed.
(ii) Determine the distance of the
8th round.
(iii) Find the total distance ran after completing all 10 rounds.
[CBSE - 2025]
Q.
How many multiples of 4 lie between 10 and 250?
Q.
If
Sn, the sum of first n terms of an AP is given by
Sn=3n2−4n, find the
nth term.
[CBSE - 2019]
Q.
Which term of the AP 3, 15, 27, 39, … will be 120 more than its 21st term?
[CBSE - 2019]
Q.
Show that
(a−b)2,(a2+b2) and (a+b)2 are in AP.
[CBSE - 2020]
Q.
If
p−1,p+1 and 2p+3 are in A.P, then the value of
p is
[CBSE - 2023]
Q.
In an A.P., if the first term
a = 7,
nth term
an=84 and the sum of first n terms
sn=22093, then
n is equal to:
[CBSE - 2024]
Q.
If the sum of first m terms of an AP is
AP is 2 m2+3 m, then its second term is:
[CBSE - 2025]
Q.
Find the 20th term from the last term of the AP: 3, 8, 13, . . ., 253.
Q.
Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.
Q.
If the terms
p−1,
p+1, and
2p+3 are consecutive terms of an Arithmetic Progression (A.P.), determine the value of
p.
Class 10 Maths Chapter 5 Questions & Answers – Arithmetic Progressions
Important Board Questions (CBSE 2019–2025)
Q1. (Easy)
If the terms (p − 1), (p + 1), and (2p + 3) are consecutive terms of an A.P., determine p.
Solution:
For consecutive terms in A.P.:
2(middle term) = first term + third term
2(p + 1) = (p − 1) + (2p + 3)
2p + 2 = 3p + 2
p = 0
Q2. (Medium)
The sum of first m terms of an A.P. is Sm = 2m2 + 3m. Find the second term of the A.P.
Solution:
First term a1 = S1 = 2(1)2 + 3(1) = 5
Second term a2 = S2 − S1
S2 = 2(2)2 + 3(2) = 8 + 6 = 14
a2 = 14 − 5 = 9
Q3. (Medium) [CBSE – 2025]
A charity run has rounds of 300 m, and each next round increases by 50 m. Total rounds = 10.
(i) Write the 4th, 5th, 6th terms.
(ii) Find the distance of the 8th round.
(iii) Find total distance after 10 rounds.
Solution:
This forms an A.P. with a = 300, d = 50.
(i) 4th term: a4 = a + 3d = 300 + 150 = 450 m
5th term: a5 = 300 + 4(50) = 500 m
6th term: a6 = 300 + 5(50) = 550 m
(ii) 8th term: a8 = a + 7d = 300 + 350 = 650 m
(iii) Total distance: S10 = 10/2 [2a + (10 − 1)d]
= 5[600 + 450] = 5 × 1050 = 5250 m
Q4. (Difficult) [CBSE – 2024]
In an A.P. of 40 terms, sum of first 9 terms is 153 and sum of last 6 terms is 687. Find first term and common difference. Also find sum of all 40 terms.
Solution:
Let first term = a, common difference = d.
Sum of first 9 terms:
S9 = 9/2[2a + 8d] = 153
⇒ 9[a + 4d] = 153
⇒ a + 4d = 17 …(1)
Last 6 terms are: a35 to a40.
a35 = a + 34d
Sum of last 6 terms:
= 6/2 [2(a + 34d) + 5d]
= 3[2a + 68d + 5d] = 3(2a + 73d) = 687
⇒ 2a + 73d = 229 …(2)
From (1): a = 17 − 4d
Put in (2): 2(17 − 4d) + 73d = 229
34 − 8d + 73d = 229
65d = 195 ⇒ d = 3
Then a = 17 − 12 = 5
Sum of 40 terms:
S40 = 40/2[2a + 39d] = 20[10 + 117] = 20 × 127 = 2540
Q5. (Medium) [CBSE – 2024]
The sum of first and eighth terms of an A.P. is 32 and their product is 60. Find a and d. Also find sum of first 20 terms.
Solution:
a1 = a, a8 = a + 7d
a + (a + 7d) = 32 ⇒ 2a + 7d = 32 …(1)
a(a + 7d) = 60 …(2)
From (1): 7d = 32 − 2a ⇒ a + 7d = 32 − a
Then (2) becomes: a(32 − a) = 60
a2 − 32a + 60 = 0
(a − 2)(a − 30) = 0 ⇒ a = 2 or 30
If a = 2, then 2(2) + 7d = 32 ⇒ 7d = 28 ⇒ d = 4
(Valid AP)
Sum of first 20 terms:
S20 = 20/2[2a + 19d] = 10[4 + 76] = 800
Answer: a = 2, d = 4, S20 = 800.
Q6. (Medium) [CBSE – 2023]
250 logs are stacked: 22 in bottom row, 21 next, 20 next and so on. Find number of rows and logs in top row.
Solution:
This is an A.P. with a = 22, d = −1, total logs Sn = 250.
Sn = n/2[2a + (n − 1)d]
250 = n/2[44 + (n − 1)(−1)]
250 = n/2[44 − n + 1] = n/2(45 − n)
500 = n(45 − n)
n2 − 45n + 500 = 0
(n − 20)(n − 25) = 0
n = 20 or 25. But top row must have positive logs:
Top row term an = 22 + (n − 1)(−1) = 23 − n
If n = 25, top = 23 − 25 = −2 (not possible)
So n = 20
Top row logs = 23 − 20 = 3
Answer: Rows = 20, top row logs = 3.
Q7. (Medium) [CBSE – 2023]
The ratio of 11th term to 17th term is 3:4. Find ratio of 5th term to 21st term. Also find ratio of sum of first 5 terms to sum of first 21 terms.
Solution:
a11 = a + 10d, a17 = a + 16d
(a + 10d)/(a + 16d) = 3/4
4a + 40d = 3a + 48d
a = 8d
a5 = a + 4d = 8d + 4d = 12d
a21 = a + 20d = 8d + 20d = 28d
So ratio a5:a21 = 12d:28d = 3:7
S5 = 5/2[2a + 4d] = 5/2[16d + 4d] = 5/2(20d) = 50d
S21 = 21/2[2a + 20d] = 21/2[16d + 20d] = 21/2(36d) = 378d
S5:S21 = 50d:378d = 25:189
Q8. (Medium) [CBSE – 2025]
If the sum of first m terms of an AP is 2m2 + 3m, then its second term is:
Solution:
Same as Q2.
S1 = 5, S2 = 14 ⇒ a2 = 9
Answer: 9
Q9. (Easy) [CBSE – 2024]
In an AP, a = 7, an = 84 and Sn = 2093. Find n.
Solution:
an = a + (n − 1)d
84 = 7 + (n − 1)d ⇒ (n − 1)d = 77 ⇒ d = 77/(n − 1) …(1)
Sn = n/2(a + an) = n/2(7 + 84) = n/2(91) = 2093
91n = 4186 ⇒ n = 46
Answer: n = 46.
Q10. (Easy) [CBSE – 2023]
If (p − 1), (p + 1) and (2p + 3) are in A.P, find p.
Solution:
Same as Q1 ⇒ p = 0
Q11. (Medium) [CBSE – 2019]
If S4 = 40 and S14 = 280 for an AP, find the sum of first n terms.
Solution:
S4 = 4/2[2a + 3d] = 2(2a + 3d) = 40 ⇒ 2a + 3d = 20 …(1)
S14 = 14/2[2a + 13d] = 7(2a + 13d) = 280 ⇒ 2a + 13d = 40 …(2)
Subtract (1) from (2): 10d = 20 ⇒ d = 2
Put in (1): 2a + 6 = 20 ⇒ a = 7
So Sn = n/2[2a + (n − 1)d]
= n/2[14 + 2n − 2] = n/2(2n + 12) = n(n + 6)
Answer: Sn = n(n + 6).
Q12. (Medium) [CBSE – 2020]
Solve: 1 + 4 + 7 + 10 + … + x = 287
Solution:
This is an AP with a = 1, d = 3, last term = x.
Let number of terms = n.
x = a + (n − 1)d = 1 + 3(n − 1) = 3n − 2 …(1)
Sum: Sn = n/2[a + x] = 287
n/2(1 + x) = 287 ⇒ n(1 + x) = 574
Put x = 3n − 2:
n(1 + 3n − 2) = 574 ⇒ n(3n − 1) = 574
3n2 − n − 574 = 0
Discriminant = 1 + 6888 = 6889 = 832
n = (1 + 83)/6 = 14
x = 3n − 2 = 3(14) − 2 = 40
Answer: x = 40.
Q13. (Medium) [CBSE – 2020]
The sum of four consecutive terms in AP is 32 and ratio of product of first and last to product of middle terms is 7:15. Find the numbers.
Solution:
Let the four consecutive terms be: a − 3d, a − d, a + d, a + 3d.
Sum: (a − 3d) + (a − d) + (a + d) + (a + 3d) = 4a = 32 ⇒ a = 8
Given ratio:
[(a − 3d)(a + 3d)] : [(a − d)(a + d)] = 7 : 15
[(a2 − 9d2</sup)] : [(a2 − d2)] = 7 : 15
Put a = 8:
(64 − 9d2)/(64 − d2) = 7/15
15(64 − 9d2) = 7(64 − d2)
960 − 135d2 = 448 − 7d2
512 = 128d2 ⇒ d2 = 4 ⇒ d = 2
Numbers: 8 − 6 = 2, 8 − 2 = 6, 8 + 2 = 10, 8 + 6 = 14
Answer: 2, 6, 10, 14.
Q14. (Medium) [CBSE – 2019]
If Sn = 3n2 − 4n, find the nth term.
Solution:
an = Sn − Sn−1
Sn−1 = 3(n − 1)2 − 4(n − 1) = 3(n2 − 2n + 1) − 4n + 4
= 3n2 − 6n + 3 − 4n + 4 = 3n2 − 10n + 7
an = (3n2 − 4n) − (3n2 − 10n + 7) = 6n − 7
Answer: an = 6n − 7.
Q15. (Medium) [CBSE – 2019]
Which term of the AP 3, 15, 27, 39, … will be 120 more than its 21st term?
Solution:
Here a = 3, d = 12
a21 = 3 + 20(12) = 243
Required term = 243 + 120 = 363
an = 3 + (n − 1)12 = 363
12(n − 1) = 360 ⇒ n − 1 = 30 ⇒ n = 31
Answer: 31st term.
Q16. (Easy) [CBSE – 2020]
Show that (a − b)2, (a2 + b2) and (a + b)2 are in AP.
Solution:
For A.P.: 2(middle) = first + third
2(a2 + b2) = (a − b)2 + (a + b)2
LHS = 2a2 + 2b2
RHS = (a2 − 2ab + b2) + (a2 + 2ab + b2) = 2a2 + 2b2
LHS = RHS, hence they are in A.P.
Q17. (Medium)
Find the 20th term from the last term of the AP: 3, 8, 13, …, 253.
Solution:
a = 3, d = 5, last term l = 253
Let number of terms be n:
l = a + (n − 1)d
253 = 3 + (n − 1)5 ⇒ 250 = 5(n − 1) ⇒ n − 1 = 50 ⇒ n = 51
20th term from last = (51 − 20 + 1)th term = 32nd term from start.
a32 = 3 + 31(5) = 3 + 155 = 158
Answer: 158.
Q18. (Medium)
Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.
Solution:
Let first term = a, common difference = d.
Third term: a + 2d = 16 …(1)
7th term − 5th term = 12
(a + 6d) − (a + 4d) = 2d = 12 ⇒ d = 6
Put d = 6 in (1): a + 12 = 16 ⇒ a = 4
Answer: AP is 4, 10, 16, 22, … (a = 4, d = 6).
Q19. (Medium)
How many multiples of 4 lie between 10 and 250?
Solution:
Multiples of 4 greater than 10 start from 12 (= 4×3).
Multiples of 4 less than 250 end at 248 (= 4×62).
So numbers are: 12, 16, 20, …, 248 (an AP)
a = 12, d = 4, last = 248
n = [(248 − 12)/4] + 1 = (236/4) + 1 = 59 + 1 = 60
Answer: 60 multiples of 4.
Q20. (Medium)
How many three-digit numbers are divisible by 7?
Solution:
Smallest 3-digit multiple of 7 is 105 (= 7×15).
Largest 3-digit multiple of 7 is 994 (= 7×142).
These form an AP: 105, 112, 119, …, 994 with d = 7
n = [(994 − 105)/7] + 1 = (889/7) + 1 = 127 + 1 = 128
Answer: 128 three-digit numbers are divisible by 7.
NCERT Solutions for Class 10 Maths Chapter 5 – FAQs
Q1. Why is Chapter 5 Arithmetic Progressions important for board exams?
Chapter 5 is frequently asked in CBSE Class 10 board exams. It includes direct formula-based questions and word problems related to sequences, making it a scoring chapter.
Q2. Which formulas are most important in Arithmetic Progressions?
- nth term: an = a + (n − 1)d
- Sum of n terms: Sn = n/2[2a + (n − 1)d]
- Also: Sn = n/2(a + l) (when last term l is known)
Q3. How to find the nth term when the sum Sn is given?
Use the relation: an = Sn − Sn−1. This method is commonly used in board questions.
Q4. What type of questions are most asked from this chapter?
CBSE usually asks word problems, finding n, finding sum of n terms, and AP formation based questions.
Q5. How should students prepare Chapter 5 for full marks?
Students should practise NCERT exercise questions, revise formulas regularly, and solve previous year CBSE questions to improve speed and accuracy.