CBSE Class 10 Maths Chapter 5 Arithmetic Progression Objective Questions

CBSE Class 10 Mathematics Chapter 5 Arithmetic Progression Objective Questions can be accessed by students to enhance their logical thinking skills, which are crucial from the exam point of view. You will find only the Multiple Choice Questions (MCQs) framed from various important topics in the chapter. Each question is provided with a detailed solution, which the CBSE students can refer to before the board examinations.

All these questions are crucial as it helps the students to prepare for CBSE Class 10 Mathematics Board Examination. As per the latest modified exam pattern, the question papers will contain more of objective-type questions. This topic is also very important from the standpoint of competitive exams. Find the CBSE Class 1o Maths Objective Questions below:

List of Sub-topics Covered in Chapter 5

5.1 General Term of AP (7 MCQs from This Topic)

5.2 Introduction to AP (6 MCQs from This Topic)

5.3 Sum of Terms in AP (7 MCQs from This Topic)

Download Free CBSE Class 10 Maths Chapter 5 – Arithmetic Progression Objective Questions PDF

General Term of AP

1. 1. Find the number of terms in each of the following APs: (i) 7, 13, 19…., 205, (ii) 18, 31/2, 13,…−47
      1. 26, 35
      2. 27, 34
      3. 35, 26
      4. 34, 27

Answer: (d) 34, 27

Solution: (i) 7, 13, 19…., 205

The first term, a = 7

The common difference, d=13−7=6

a_n=205

Using formula an=a+ (n−1) d to find the n^th term of arithmetic progression, we get

205=7+ (n−1)6

⇒205=6n+1

⇒204=6n

⇒n=204/6=34

Therefore, there are 34 terms in the given arithmetic progression.

(ii) 18, 31/2, 13,…−47

The first term, a=18

Common difference, d= (31/2) −18=−5/2

a_n = − 47

Using formula a_n=a+ (n−1) d to find the n^th term of arithmetic progression, we get

−47=18 + (n−1) (−5/2)

⇒−94=36−5n+5

⇒5n=135

⇒n=135/5=27

Therefore, there are 27 terms in the given arithmetic progression.

2. 2. Find the 31st term of an AP whose 11th term is 38 and 16th term is 73.
      1. 185
      2. 210
      3. 178
      4. 150

Answer: (c) 178

Solution: We are given that a₁₁=38 and a₁₆=73 where a₁₁ is the 11^th term and  a₁₆ is the 16^th term of an AP.

Using formula a_n=a+ (n−1) d to find the nth term of arithmetic progression, we get

38=a+10d….. (i)

73=a+15d …… (ii)

equation (ii) – equation (i) gives,

35 = 5d

d = 7….. (iii)

Substituting (iii) in (i), we get a = -32

a₃₁=−32+(31−1)(7)

⇒−32+210=178

Therefore, the 31^st term of AP is 178.

3. 3. If the third and the ninth terms of an AP are 4 and -8, respectively, which term of this AP is zero?
      1. 5^th
      2. 4^th
      3. 3^rd
      4. 6^th

Answer: (a) 5^th

Solution: It is given that the 3rd and 9th terms of AP are 4 and -8, respectively.

It means a₃=4 and a₉=−8

Where a₃ and a₉ are the third and ninth terms, respectively.

Using formula a_n=a+(n−1)d to find the n^th term of arithmetic progression, we get

4=a+(3−1)d

⇒4 = a+2d…(i)

−8 = a+(9−1)d

⇒−8=a+8d…(ii)

From equation (i), we have  a=4−2d

Substituting in equation (ii), we get

−8=4−2d+8d

⇒−12 = 6d

⇒d =−12/6 =−2

Solving for (a), we get

⇒−8 = a−16

⇒a = 8

Therefore, first term a = 8 and common difference d =−2

We know  a_n = a+ (n−1) d (where a_n is the n^th term)

Finding the value of n where an=0

0=8+ (n−1) (−2)

⇒0 = 8−2n+2

⇒0 = 10−2n

⇒2n = 10

⇒n = 10/2=5

Therefore, the 5th term is equal to 0.

4. 4. Which term of the AP: 3, 15, 27, 39, … will be 132 more than its 54^th term?
      1. 70
      2. 65
      3. 80
      4. 55

Answer: (b) 65

Solution: Let’s first calculate the 54^th of the given AP.

First term = a = 3

Common difference = d =15 – 3 = 12

Using formula a_n=a+ (n−1) d,   to find the n^th  term of arithmetic progression, we get

a₅₄=a+ (54−1) d

a₅₄ = 3 + 53 (12) =3 + 636=639

We want to find which term is 132 more than its 54^th term. Let’s suppose it is the n^th term which is 132 more than the 54^th term.

Therefore, we can say that

a_n = a₅₄+132

a_n=a+ (n−1) d=3+ (n−1) (12)}

⇒3+ (n−1)12=639+132

⇒3+12n−12=771

⇒12n−9=771

⇒12n=780

⇒n= 780/12=65

Therefore, the 65^th term is 132 more than its 54^th term.

5. 5. Two APs have the same common difference. The difference between their 100^th terms is 100; what is the difference between their 1000^th terms?
      1. 200
      2. 150
      3. 100
      4. 55

Answer: (c) 100

Solution: Let the first term of the first AP = a

Let the first term of the 2nd AP = a′

It is given that their common difference is the same. Let their common difference be d.

It is given that difference between their 100^th terms is 100. Using formula a_n = a + (n−1) d,  to find the n^th term of arithmetic progression, we can say that

a+ (100−1) d− (a′ +(100−1)d)=a+99d−a′−99d=100

⇒a−a′=100      …….. (1)

We want to find the difference between their 1000^th terms, which means we want to calculate the following:

a+ (1000−1) d−(a′+(1000−1)d)=a+999d−a′−999d=a−a′

Putting equation (1) in the above equation, we get,

a+ (1000−1) d−(a′+(1000−1)d)=a+999d−a′-999d=a−a′=100

Therefore, the difference between their 1000^th terms would be equal to 100

6. 6. How many three-digit numbers are divisible by 7?
      1. 112
      2. 114
      3. 128
      4. 110

Answer: (c) 128

Solution: We have an AP starting at 105 because it is the first three-digit number divisible by 7.

AP will end at 994 because it is the last three-digit number divisible by 7.

Therefore, we have an AP 105,112,119…994

The first term, a = 105

Common difference, d = 112 – 105 = 7

Using formula a_n=a+ (n−1) d to find the n^th term of arithmetic progression, we can say that

994=105+ (n−1) (7)

⇒994=105+ (n−1) (7)

⇒889=7(n−1)

⇒n−1=889/7⇒n=127+1=128

994 is the 128^th term of AP. Therefore, there are 128 terms in AP. In other words, we can also say that there are 128 three-digit numbers divisible by 7.

7. 7. For what value of n are the n^th terms of two AP’s: 63, 65, 67 …. and 3, 10, 17,.. equal?
      1. 11
      2. 14
      3. 12
      4. 13

Answer: (d) 13

Solution: Let’s first consider AP: 63, 65, 67…..

First term =a=63

Common difference =d=65−63=2

Using formula a_n=a+ (n−1) d,   to find the n^th term of arithmetic progression, we can say that

a_n=63+ (n−1) (2)                      (1)

Now, consider second AP 3, 10, 17…

First term = a = 3

Common difference = d = 10 -3 = 7

Using formula a_n=a+ (n−1) d,   to find the n^th term of arithmetic progression, we can say that

a_n=3+(n−1)(7)                    (2)

According to the given condition, we can write

(1) = (2)

⇒63+ (n−1) (2) =3+ (n−1) (7)

⇒63+2n−2=3+7n−7

⇒65=5n

⇒n=65/5=13

Therefore, the 13th terms of both APs are equal.

Introduction to AP

8. 8. If (x+1), 3x and (4x+2) are the first three terms of an AP, then its 5th term is:
      1. 14
      2. 19
      3. 24
      4. 28

Answer: (d) 28

Solution: Given (x+1), 3x, (4x+2) are in AP.

Hence, the difference between two consecutive terms will be the same.

Hence, 3x–(x+1) = (4x+2)–3x

⇒2x−1=x+2

⇒x=3

So, the first term,

a=(x+1) =4.

The common difference,

d=9–4=5.

The n^th term of an AP is given by

t_n =a+ (n−1) d

⇒t₅=4+4(5) =24.

9. 9. The sum of the first ten terms of an A.P. is four times the sum of its first five terms, then the ratio of the first term and the common difference is
      1. ½
      2. ¼
      3. 4
      4. 1

Answer: (a) ½

Solution: Let S₁₀ be the sum of the first 10 terms and S5 be the sum of the first 5 terms.

We know,

S_{n = (n/2) [2a + (n-1) d]}

Given,

S₁₀=4S₅

⇒ (10/2) [2a + (10-1) d] = 4 × (5/2) [2a + (5-1) d]

⇒ (10/2) [2a+ 9d] = 4 × (5/2) [2a+ 4d]

⇒ 2a+ 9d = 4a+ 8d

⇒ a/d= ½

10. 10. Find the common difference (d) in the following APs, respectively.

(i)     20, 40, 60, 80, 100,…

(ii)    5, 0, -5, -10, -15, …

(iii)   2, 2, 2, 2, 2..

10. 10. 1. 20, 5, 0
        2. -20, -5, 0
        3. -20, 5, 0
        4. 20, -5, 0

Answer: (d) 20, -5, 0

Solution: The common difference is

(i)  20, 40, 60, 80, 100,

d= 40-20 =20

(ii) 5, 0, -5, -10, -15, …

d= 0-5 = -5

(iii) 2, 2, 2, 2, 2….

d= 2-2 =0

11. 11. Which of the following sequences form an AP?

(a) 2, 4, 8, 16……..

(b) 2, 3, 5, 7, 11…….

(c) -1, -1.25, -1.5, -1.75……

(d) 1, -1, -3, -5, -7…………

Answer: (ic) -1, -1.25, -1.5, -1.75…… and

(d) 1, -1, -3, -5, -7…………

Solution: Consider each list of numbers:

(a) 2, 4, 8, 16……..

Difference between the first two terms = 4 – 2 = 2

Difference between the third and second term = 8 – 4 = 4

Since a₂–a₁≠a₃–a₂, this sequence is not an AP

(b) 2, 3, 5, 7, 11……..

Difference between the first two terms = 3 – 2 = 1

Difference between the third and second term = 5 – 3 = 2

Since a₂–a₁≠a₃–a₂, this sequence is not an AP

(c) -1, -1.25, -1.5, -1.75…….

Difference between the first two terms = -1.25 – (-1) = -0.25

Difference between the third and second term = -1.5 – (-1.25) = -0.25

Difference between the fourth and third term = -1.75 – (-1.5) = -0.25

Since a₂–a₁=a₃−a₂=a₄−a₃, this sequence is an AP

(d) 1, -1, -3, -5, -7…………

Difference between the first two terms = -1 – 1 = -2

Difference between the third and second term = -3 – (-1) = -2

Difference between the fourth and third term = -5 – (-3) = -2

Since a₂–a₁=a₃−a₂=a₄−a₃, this sequence is an AP

12. 12. What are the conditions for a sequence to be an AP?
        1. The sum of two consecutive terms should be constant.
        2. The product of two consecutive numbers should be constant.
        3. The difference between two consecutive terms should be constant.
        4. The ratio of two consecutive terms should be constant.

Answer: (c) The difference between two consecutive terms should be constant.

Solution: Let a1, a₂, a₃, a₄, a₅, a_6, a_7, a₈… be a sequence.

For this sequence to be an AP, the difference between any two consecutive terms should be constant.

a₂–a₁=a₃–a₂=a₄–a₃=d

This difference is called the common difference of the AP and is denoted by d.

So an AP can also be represented in this form as well a, a+d, a+2d…

13. 13. Which of the following list of numbers forms an AP?

(a) 4, 4 + √3 , 4 + 2√3, 4 + 3√3, 4 + 4√3

(b) 0.3, 0.33, 0.333, 0.3333, 0.33333

(c) 3/5, 6/5, 9/5, 12/5, 3

(d) -1/5, -1/5, -1/5, -1/5, -1/5

Answer: (a), (c) and (d)

Solution: Consider each series (a) 4, 4 + √3, 4 + 2 √3, 4 + 3 √3, 4 + 4 √3

Difference between first two consecutive terms = 4 + √3 – 4 = √3

Difference between third and second consecutive terms = 4 + 2 √3 – 4 + √3 =  √3

Difference between fourth and third consecutive terms = 4 + 3 √3– 4 + 2 √3 =  √3

Since a₂–a₁ = a₃−a₂ = a₄−a₃

This series is an AP

(b) 0.3, 0.33, 0.333, 0.3333, 0.33333

Difference between first two consecutive terms = 0.33-0.3 = 0.03

Difference between third and second consecutive terms = 0.333 – 0.33 = 0.003

Since  a₂–a₁ ≠ a₃−a₂

This series is not an AP

(c) 3/5, 6/5, 9/5, 12/5, 3

Difference between first two consecutive terms = (6/5)–(3/5) = 3/5

Difference between third and second consecutive terms = (9/5)–(6/5) = 3/5

Difference between fourth and third consecutive terms = (12/5) – (9/5) = 3/5

Since  a₂–a₁ = a₃–a₂ = a₄–a₃

This series is an AP

(d) -1/5, -1/5, -1/5, -1/5, -1/5

Difference between first two consecutive terms = -1/5 – (- 1/5) = 0

Difference between third and second consecutive terms = -1/5 – (-1/5) = 0

Difference between fourth and third consecutive terms = -1/5 – (-1/5) = 0

Since a₂–a₁ = a₃–a₂ = a₄–a₃

This series is an AP.

Sum of Terms in AP

14. 14. The first term of an AP is 5, the last term is 50, and the sum is 440. Find the number of terms and the common difference.
        1. 17; 3
        2. 16; 2
        3. 17; 2
        4. 16; 3

Answer: (d) 16; 3

Solution: First term, a=5

Last term, l=50

S_n=440

Applying the formula S_n= n/2 (a+l) to find the sum of n terms of AP, we get

440= n/2 (5+50)

⇒440/55=n/2

⇒8=n/ 2

⇒n=16

Applying the formula, S_n=n/2(2a+ (n−1) d) to find the sum of n terms of AP and putting the value of n, we get

440=16/2(2(5) + (16−1) d)

⇒440=8(10+15d)

⇒10+15d=55

⇒15d=45

⇒d=45/15=3

15. 15. The first and the last terms of an AP are 17 and 350, respectively. If the common difference is 9, how many terms are there, and what is their sum?
        1. 38; 7114
        2. 32; 7114
        3. 38; 6973
        4. 32; 6973

Answer: (c) 38; 6973

Solution: First term = a = 17

Last term = l = 350

Common difference = d = 9

Using formula a_n = a + (n−1) d to find the nth term of arithmetic progression, we can say that

350=17+ (n−1) (9)

⇒350=17+9n−9

⇒342=9n

⇒n=342/9=38

Applying the formula, S_n=n/2(2a+ (n−1) d) to find the sum of n terms of AP and putting the value of n, we get

S₃₈=38/2(34+ (38−1) (9))

⇒S₃₈=19(34+333) =6973

Therefore, there are 38 terms, and their sum is equal to 6973.

16. 16. Find the sum of the first 22 terms of an AP in which d = 7 and the 22^nd term is 149.
        1. 1623
        2. 1712
        3. 1542
        4. 1661

Answer: (d) 1661

Solution: It is given that the 22nd term is equal to 149.

It means a₂₂=149

Using formula a_n=a+ (n−1) d,   to find the n^th term of AP, we can say that

149=a+ (22−1) (7)

⇒149=a+147

⇒a=2

Applying the formula, S_n = n/2(2a+ (n−1) d) to find the sum of n terms of AP and putting the value of a, we get

S₂₂= 22/2(4+ (22−1) (7))

⇒S₂₂=11(4+147)

⇒S₂₂=1661

Therefore, the sum of the first 22 terms of AP is equal to 1661.

17. 17. Find the sum of the first 51 terms of an AP whose second and third terms are 14 and 18, respectively.
        1. 5610
        2. 5840
        3. 5320
        4. 5000

Answer: (a) 5610

Solution: It is given that the second and third terms of AP are 14 and 18, respectively.

Using formula a_n=a+ (n−1) d,   to find the n^th term of arithmetic progression, we can say that

14=a+ (2−1) d

⇒14=a+ d                                   (1)

And,    18=a+ (3−1) d

⇒18=a+2d                                 (2)

These are equations consisting of two variables. We can solve them by the method of substitution.

Using equation (1), we can say that a=14−d

Putting the value of a in equation (2), we can say that

18=14−d+2d

⇒d=4

Therefore, the common difference d=4

Putting the value of d in equation number (1), we can say that

18=a+2(4)

⇒a=10

Applying the formula, S_n= n/2(2a+ (n−1) d) to find the sum of n terms of AP, we get

S₁₅=51/2(20+ (51−1) 4) =51/2(20+200) =51×110 =5610

Therefore, the sum of the first 51 terms of an AP is equal to 5610.

18. 18. The sum of four consecutive numbers in an A.P. with d > 0 is 20. The sum of their square is 120, then the middle terms are
        1. 8, 10
        2. 6, 8
        3. 4, 6
        4. 2, 4

Answer: (c) 4, 6

Solution: Let the numbers be a−3d,a−d,a+d,a+3d

Given: a−3d+a—d+a+d+a+3d=20

4a=20

a=5 and

(a−3d)²+ (a−d)² + (a+3d)² + (a+3d)²=120

(4)(a)² + 20 d²=120

(4)(5)²+20 d²=120

d²=1

d=+1 or−1

Hence, the numbers are 2, 4, 6, and 8

19. 19. Find the sum of all the non-negative terms of the following sequence: 100, 97, 94, ….
        1. 1717
        2. 1719
        3. 1721
        4. 1723

Answer: (a) 1717

Solution: Here, t₁ = 100, common difference, d = t₂ – t₁ = 97 – 100 = -3

t_n = t₁ + (n-1) d ⇒100 + (n-1)(-3) = 100 – 3(n-1) = 103 – 3n.

Let t_n be the first negative term. i.e.,

t_n < 0 ⇒ 103 – 3n < 0 ⇒ n > 103/3 > 34

That is, the 35^th term will be negative.

The sum of the first 34 terms = (34/2) (first term + last term)

= 17(100+100+ (34−1) (−3)) =1717

20. 20. What is the sum of all 3 digit numbers that leave a remainder of ‘2’ when divided by 3?
        1. 149700
        2. 164749
        3. 164850
        4. 897

Answer: (c) 164850

Solution: The smallest 3-digit number that will leave a remainder of 2 when divided by 3 is 101.

The next number that will leave a remainder of 2 when divided by 3 is 104, 107,…

The largest 3-digit number that will leave a remainder of 2 when divided by 3 is 998.

So, it is an AP with the first term being 101 and the last term being 998, and the common difference being 3.

Sum of an AP =  n/2(a+l)

we know that in an A.P., the nth term a_n=a₁+ (n−1)×d

In this case, therefore,

998 = 101 + (n – 1) × 3

⇒897= (n−1) ×3

⇒299= (n−1)

⇒n=300

Therefore, the sum of the AP will be (101+198)/2 ×300= 164850

CBSE Class 10 Maths Chapter 5 Extra MCQs

1. In an AP, if d = –4, n = 7, a_n = 4. Calculate a.
(a) 20
(b) 7
(c) 28
(d) 6

2. What is the sum of the first five multiples of 3?
(a)75
(b) 55
(c) 65
(d) 45

In this chapter, the concepts discussed include patterns in succeeding terms obtained by adding a fixed number to the preceding terms. Students will understand how to find nth terms and the sum of n consecutive terms. Also, they will learn arithmetic progression effectively when they solve daily life problems.

This chapter has Arithmetic Progression Derivation of the nth term and sum of the first n terms of an A.P. and their application in solving daily life problems.

Keep visiting BYJU’S to get complete assistance for CBSE Class 10 Board exams. At BYJU’S, students can obtain several sample papers, question papers, notes, textbooks, videos, animations and effective preparation tips, which can help them to score good grades in the Class 10 board exams.