Arithmetic Progressions Chapter 5: NCERT Solutions for Class 10

The ncert exemplar class 10 maths arithmetic progression is provided here in PDF format free for students to download. They can practice these questions and ace them to prepare for the CBSE exam. These materials are prepared by our subject experts, keeping into consideration the CBSE (2024-25). Students can download exemplar problems with solutions and use them as per their convenience with learning about new topics and also find answers to all questions given in the chapter. Furthermore, students can use this as a reference tool to do a complete revision of the entire chapter before the board exam. Simply Acad mainly focuses upon the prior concern for the students and hence provides all the solutions at one go at a platform for free and for a 24/7 access to the same.

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NCERT Exemplar Class 10 – Arithmetic Progressions Question 1 to 10

Question 1.

For the AP, d=–4d = –4d=–4, n=7n = 7n=7, an=4a_n = 4an​=4, find aaa:

Options:

• (A) 6
• (B) 7
• (C) 20
• (D) 28

Answer 1: (D) 28

Question 2.

For the AP, a=3.5a = 3.5a=3.5, d=0d = 0d=0, n=101n = 101n=101, find ana_nan​:

Options:

• (A) 0
• (B) 3.5
• (C) 103.5
• (D) 104.5

Answer 2: (B) 3.5

Question 3.

The list of numbers –10, –6, –2, 2,… is:

Options:

• (A) an A.P. with d=–16d = –16d=–16
• (B) an A.P. with d=4d = 4d=4
• (C) an A.P. with d=–4d = –4d=–4
• (D) not an A.P.

Answer 3: (B) an A.P. with d=4d = 4d=4

Question 4.

The 11th term of the AP: –5, –52\frac{5}{2}25​, 0, 52\frac{5}{2}25​, … is:

Options:

• (A) –20
• (B) 20
• (C) –30
• (D) 30

Answer 4: (B) 20

Question 5.

The first four terms of the AP, whose first term is –2 and their common difference is –2, are:

Options:

• (A) –2, 0, 2, 4
• (B) –2, 4, –8, 16
• (C) –2, –4, –6, –8
• (D) –2, –4, –8, –16

Answer 5: (C) –2, –4, –6, –8

Question 6.

The 21st term of the AP, whose first two terms are –3 and 4, is:

Options:

• (A) 17
• (B) 137
• (C) 143
• (D) –143

Answer 6: (B) 137

Question 7.

If the 2nd term of the AP is 13 and the 5th term is 25, then what is the 7th term?

Options:

• (A) 30
• (B) 33
• (C) 37
• (D) 38

Answer 7: (B) 33

Question 8.

Which term of the AP: 21, 42, 63, 84… is 210?

Options:

• (A) 9th
• (B) 10th
• (C) 11th
• (D) 12th

Answer 8: (B) 10th

Question 9.

If the common difference of the AP is 5, what is a18–a13a_{18} – a_{13}a18​–a13​?

Options:

• (A) 5
• (B) 20
• (C) 25
• (D) 30

Answer 9: (C) 25

Question 10.

What is the common difference of the AP for which a18–a14=32a_{18} – a_{14} = 32a18​–a14​=32?

Options:

• (A) 8
• (B) –8
• (C) –4
• (D) 4

Answer 10: (A) 8

NCERT Exemplar Class 10 – Arithmetic Progressions Question 11 to 20

Question 11.

Two APs have the same common difference. The first term for one of these is –1, and that for the other is –8. Then the difference for the 4th term is:

Options:

• (A) –1
• (B) –8
• (C) 7
• (D) –9

Answer 11: (C) 7

Question 12.

If 7 times the 7th term of the AP is equal to 11 times the 11th term, the 18th term will be:

Options:

• (A) 7
• (B) 11
• (C) 18
• (D) 0

Answer 12: (D) 0

Question 13.

The 4th term from the end of the AP: –11, –8, –5, …, 49 is:

Options:

• (A) 37
• (B) 40
• (C) 43
• (D) 58

Answer 13: (B) 40

Question 14.

The famous mathematician associated with finding the sum of the first 100 natural numbers is:

Options:

• (A) Pythagoras
• (B) Newton
• (C) Gauss
• (D) Euclid

Answer 14: (C) Gauss

Question 15.

If the first term of the AP is –5 and the common difference is 2, the sum of the first 6 terms is:

Options:

• (A) 0
• (B) 5
• (C) 6
• (D) 15

Answer 15: (A) 0

Question 16.

The sum of the first 16 terms of the AP: 10, 6, 2,… is:

Options:

• (A) –320
• (B) 320
• (C) –352
• (D) –400

Answer 16: (A) –320

Question 17.

For the AP a=1a = 1a=1, an=20a_n = 20an​=20, and Sn=399S_n = 399Sn​=399, find nnn:

Options:

• (A) 19
• (B) 21
• (C) 38
• (D) 42

Answer 17: (C) 38

Question 18.

The sum of the first five multiples of 3 is:

Options:

• (A) 45
• (B) 55
• (C) 65
• (D) 75

Answer 18: (A) 45

Question 19.

Which among the following forms an AP?

Answer 19:

• (i) –1, –1, –1, –1,… (Yes, forms an AP)
• (ii) 0, 2, 0, 2,… (No, does not form an AP)
• (iii) 1, 1, 2, 2, 3, 3… (No, does not form an AP)
• (iv) 11, 22, 33… (Yes, forms an AP)
• (v) 1/2, 1/3, 1/4, … (No, does not form an AP)
• (vi) 2, 22, 23, 24, … (No, does not form an AP)
• (vii) 3\sqrt{3}3​, 12\sqrt{12}12​, 27\sqrt{27}27​, 48\sqrt{48}48​, … (Yes, forms an AP)

Question 20.

State whether it is true that –1, −32-\frac{3}{2}−23​, –2, 52\frac{5}{2}25​, … forms an AP as a2–a1=a3–a2a_2 – a_1 = a_3 – a_2a2​–a1​=a3​–a2​:

Answer 20: False

NCERT Exemplar Class 10 – Arithmetic Progressions Question 11 to 20

Question 21.

For the AP: –3, –7, –11, …, could we directly find a30–a20a_{30} – a_{20}a30​–a20​ without actually finding a30a_{30}a30​ and a20a_{20}a20​? Justify your answer.

Answer 21: Yes, a30–a20=–40a_{30} – a_{20} = –40a30​–a20​=–40

Question 22.

Verify that each of the following is an AP, and write the next three terms:

Answer 22:

• (i) 0, 14\frac{1}{4}41​, 12\frac{1}{2}21​, 34\frac{3}{4}43​,… (Yes, next three terms: 1, 54\frac{5}{4}45​, 32\frac{3}{2}23​)
• (ii) 5, 143\frac{14}{3}314​, 133\frac{13}{3}313​, 4… (Yes, next three terms: 113\frac{11}{3}311​, 103\frac{10}{3}310​, 3)
• (iii) 3\sqrt{3}3​, 232\sqrt{3}23​, 333\sqrt{3}33​,… (Yes, next three terms: 434\sqrt{3}43​, 535\sqrt{3}53​, 636\sqrt{3}63​)
• (iv) a+ba + ba+b, (a+1)+b(a + 1) + b(a+1)+b, (a+1)+(b+1)(a + 1) + (b + 1)(a+1)+(b+1), … (Yes, next three terms: (a+2)+(b+1)(a + 2) + (b + 1)(a+2)+(b+1), (a+2)+(b+2)(a + 2) + (b + 2)(a+2)+(b+2), (a+3)+(b+2)(a + 3) + (b + 2)(a+3)+(b+2))
• (v) aaa, 2a+12a + 12a+1, 3a+23a + 23a+2, 4a+34a + 34a+3,… (Yes, next three terms: 5a+45a + 45a+4, 6a+56a + 56a+5, 7a+67a + 67a+6)

Question 23.

Write the first four terms of the AP if the first term aaa and the common difference ddd are given as:

Answer 23:

• (i) a=10a = 10a=10, d=10d = 10d=10: 10, 20, 30, 40
• (ii) a=−2a = -2a=−2, d=0d = 0d=0: –2, –2, –2, –2
• (iii) a=4a = 4a=4, d=–3d = –3d=–3: 4, 1, –2, –5
• (iv) a=−1a = -1a=−1, d=12d = \frac{1}{2}d=21​: –1, −12-\frac{1}{2}−21​, 0, 12\frac{1}{2}21​
• (v) a=−1.25a = -1.25a=−1.25, d=−0.25d = -0.25d=−0.25: –1.25, –1.50, –1.75, –2.00

Question 24.

For the following APs, write the first term and the common difference:

Answer 24:

• (i) 3, 1, –1, –3 … a=3a = 3a=3, d=−2d = -2d=−2
• (ii) -5, –1, 3, 7 … a=−5a = -5a=−5, d=4d = 4d=4
• (iii) 13\frac{1}{3}31​, 53\frac{5}{3}35​, 93\frac{9}{3}39​, 133\frac{13}{3}313​ … a=13a = \frac{1}{3}a=31​, d=43d = \frac{4}{3}d=34​
• (iv) 0.6, 1.7, 2.8, 3.9 … a=0.6a = 0.6a=0.6, d=1.1d = 1.1d=1.1

Question 25.

Which among the following are APs? Find the common difference ddd and write three more terms:

Answer 25:

• (i) 2, 4, 8, 16 … (Not an AP)
• (ii) 2, 52\frac{5}{2}25​, 3, 72\frac{7}{2}27​ … (Yes, d=12d = \frac{1}{2}d=21​, next three terms: 4, 92\frac{9}{2}29​, 5)
• (iii) -1.2, -3.2, -5.2, -7.2 … (Yes, d=−2d = -2d=−2, next three terms: -9.2, -11.2, -13.2)
• (iv) -10, –6, –2, 2 … (Yes, d=4d = 4d=4, next three terms: 6, 10, 14)
• (v) 3, 3+23 + \sqrt{2}3+2​, 3+223 + 2\sqrt{2}3+22​, 3+323 + 3\sqrt{2}3+32​ … (Yes, d=2d = \sqrt{2}d=2​, next three terms: 3+423 + 4\sqrt{2}3+42​, 3+523 + 5\sqrt{2}3+52​, 3+623 + 6\sqrt{2}3+62​)
• (vi) 0.2, 0.22, 0.222, 0.2222 … (Not an AP)
• (vii) 0, -4, -8, -12 … (Yes, d=−4d = -4d=−4, next three terms: -16, -20, -24)
• (viii) −12-\frac{1}{2}−21​, −12-\frac{1}{2}−21​, −12-\frac{1}{2}−21​, −12-\frac{1}{2}−21​ … (Yes, d=0d = 0d=0, next three terms: −12-\frac{1}{2}−21​, −12-\frac{1}{2}−21​, −12-\frac{1}{2}−21​)
• (ix) 1, 3, 9, 27 … (Not an AP)
• (x) aaa, 2a2a2a, 3a3a3a, 4a4a4a … (Yes, d=ad = ad=a, next three terms: 5a5a5a, 6a6a6a, 7a7a7a)
• (xi) aaa, a2a^2a2, a3a^3a3, a4a^4a4… (Not an AP)
• (xii) 2\sqrt{2}2​, 8\sqrt{8}8​, 18\sqrt{18}18​, 32\sqrt{32}32​ … (Yes, d=2d = \sqrt{2}d=2​, next three terms: 50\sqrt{50}50​, 72\sqrt{72}72​, 98\sqrt{98}98​)
• (xiii) 3\sqrt{3}3​, 6\sqrt{6}6​, 9\sqrt{9}9​, 12\sqrt{12}12​ … (Not an AP)
• (xiv) 12, 32, 52, 72 … (Not an AP)
• (xv) 12, 52, 72, 73 … (Yes, d=24d = 24d=24, next three terms: 97, 121, 145)

Question 26.

Write the first three terms of the APs if aaa and ddd are given below as:

Answer 26:

• (i) a=12a = \frac{1}{2}a=21​, d=−16d = -\frac{1}{6}d=−61​: 12\frac{1}{2}21​, 13\frac{1}{3}31​, 16\frac{1}{6}61​
• (ii) a=−5a = -5a=−5, d=−3d = -3d=−3: -5, -8, -11
• (iii) a=2a = \sqrt{2}a=2​, d=12d = \frac{1}{\sqrt{2}}d=2​1​: 2\sqrt{2}2​, 32\frac{3}{\sqrt{2}}2​3​, 42\frac{4}{\sqrt{2}}2​4​

Question 27.

Find aaa, bbb, and ccc so that the following numbers are of the AP: aaa, 7, bbb, 23, ccc:

Answer 27:

• a=−1a = -1a=−1
• b=15b = 15b=15
• c=31c = 31c=31

Question 28.

Determine the AP in which the fifth term is 19 and the difference of the eighth term from the thirteenth term is 20:

Answer 28: AP: 3, 7, 11, …

Question 29.

The eighth term of the AP is half its second term, and the eleventh term exceeds one-third of the fourth term by 1. Find the 15th term:

Answer 29: 15th term: 3

Question 30.

An AP contains 37 terms. The sum of the three middle-most terms is 225, and the sum of the last three is 429. Find the AP:

Answer 30: AP: 3, 7, 11, …

NCERT Exemplar Class 10 – Arithmetic Progressions Question 31 to 40

Question 31.

Check if –150 is a term of the AP 11, 8, 5, 2, …:

Answer 31: No, –150 is not a term of this AP.

Question 32.

Find the 31st term of the AP whose 11th term is 38 and the 16th term is 73:

Answer 32: 31st term: 178

Question 33.

An AP consists of 50 terms, of which the 3rd term is 12 and the last term is 106. Find the 29th term:

Answer 33: 29th term: 64

Question 34.

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

Answer 34: The 5th term is zero.

Question 35.

If the 17th term of the AP exceeds its 10th term by 7, find the common difference:

Answer 35: Common difference: 1

Question 36.

Which term of the AP 3, 15, 27, 39,… will be 132 more than the 54th term?

Answer 36: 65th term

Question 37.

The sum of the first five terms of the AP and the sum of the first seven terms for the same AP is 167. If the sum of the first ten terms of the AP is 235, find the sum of the first twenty terms:

Answer 37: Sum of the first 20 terms: 970

Question 38.

Find the sum of the integers between 1 and 500 that are divisible by 2 and 5:

Answer 38:

• (i) S=12,250S = 12,250S=12,250
• (ii) S=12,750S = 12,750S=12,750
• (iii) S=75,250S = 75,250S=75,250

Question 39.

Two APs have equal common differences. The difference for the 100th term is 100, what is the difference for the 1000th terms?

Answer 39: Difference: 100

Question 40.

How many three-digit numbers are divisible by 7?

Answer 40: 128 numbers

NCERT Exemplar Class 10 – Arithmetic Progressions Question 41 to 47

Question 41.

How many multiples of 4 lie between 10 and 250?

Answer 41: 60 multiples

Question 42.

For what value of nnn are the nth terms of the two APs 63,65,67,…63, 65, 67,…63,65,67,… and 3,10,17,…3, 10, 17,…3,10,17,… the same?

Answer 42: n=13n = 13n=13

Question 43.

Find the AP whose third term is 16 and the 7th term exceeds the 5th term by 12:

Answer 43: AP: 4, 10, 16, 22, …

Question 44.

Find the 20th term from the last term of the AP 3,8,13,…,2533, 8, 13, …, 2533,8,13,…,253:

Answer 44: 20th term: 158

Question 45.

The sum of the 4th and 8th terms of the AP is 24, and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP:

Answer 45: First three terms: –13, –8, –3

Question 46.

Subba Rao started his work in 1995 at an annual salary of Rs 5000 and received an increment of Rs 200 each year. In which year did his income reach Rs 7000?

Answer 46: In the 11th year, his salary reached Rs 7000.

Question 47.

Ramkali saved Rs 5 in the first week of the year and increased her weekly saving by Rs 1.75. If in the nth week, her weekly savings become Rs 20.75, find nnn:

Answer 47: n=10n = 10n=10