 # CBSE Class 10 Maths Chapter 5- Arithmetic Progression Objective Questions

Here you get the CBSE Class 10 Mathematics chapter 5- Arithmetic Progression Objective Questions. You will find only the Multiple Choice Questions (MCQs) framed from various important topics in the chapter. Each question is also provided with a detailed solution, which the students can refer to at a later stage.

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 This Chapter

5.1 General Term of AP (7 MCQs From The Topic)

5.2 Introduction to AP (6 MCQs Listed From The Topic)

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

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

### General Term of AP

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

Solution: (i) 7, 13, 19…., 205

First term, a=7

Common difference, d=13−7=6

an=205

Using formula an=a+ (n−1) d to find nth 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

First term, a=18

Common difference, d= (31/2) −18=−5/2

an = − 47

Using formula an=a+ (n−1) d to find nth 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.

1. 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

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

Using formula an=a+ (n−1) d to find 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, 31st term of AP is 178.

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

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

It means a3=4 and a9=−8

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

Using formula an=a+(n−1)d to find nth 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 have

−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  an = a+ (n−1) d (where an is the nth term)

Finding 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, 5th term is equal to 0.

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

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

First term = a = 3

Common difference = d =15 – 3 = 12

Using formula an=a+ (n−1) d,   to find nth  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 54th term. Let’s suppose it is nth term which is 132 more than 54th term.

Therefore, we can say that

an = a54+132

an=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, 65th term is 132 more than its 54th term.

1. Two AP’s have the same common difference. The difference between their 100th terms 100, what is the difference between their 1000th terms.
1. 200
2. 150
3. 100
4. 55

Solution: Let first term of first AP = a

Let first term of 2nd AP = a′

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

It is given that difference between their 100th terms is 100. Using formula an = a + (n−1) d,  to find nth 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 difference between their 1000th terms which means we want to calculate:

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

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

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

First term, a = 105

Common difference, d = 112 – 105 = 7

Using formula an=a+ (n−1) d, to find  nth 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 128th 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.

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

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

First term =a=63

Common difference =d=65−63=2

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

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

Now, consider second AP 3, 10, 17…

First term = a = 3

Common difference = d = 10 -3 = 7

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

an=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, 13th terms of both the AP’s are equal.

### Introduction to AP

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

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

Hence, the difference of two consecutive terms will be 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 nth term of an AP is given by

tn =a+ (n−1) d

⇒t5=4+4(5) =24.

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

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

We know,

Sn = (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

1. 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..

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

Solution: 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

1. Which of the following sequences form an AP?

(i) 2, 4, 8, 16……..

(ii) 2, 3, 5, 7, 11…….

(iii) -1, -1.25, -1.5, -1.75……

(iv) 1, -1, -3, -5, -7…………

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

(iv) 1, -1, -3, -5, -7…………

Solution: Consider each list of numbers:

(i) 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

(ii) 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

(iii) -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

(iv) 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

1. 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, a6, a7, 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…

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

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

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

(iii) 3/5, 6/5, 9/5, 12/5, 3

(iv) -1/5, -1/5, -1/5, -1/5, -1/5

Solution: Consider each series (i) 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

(ii) 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

(iii) 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

(iv) -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

1. 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

Solution: First term, a=5

Last term, l=50

Sn=440

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

440= n/2 (5+50)

⇒440/55=n/2

⇒8=n/ 2

⇒n=16

Applying formula, Sn=n/2(2a+ (n−1) d) to find sum of n terms of AP and putting 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

1. 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

Solution: First term = a = 17

Last term = l = 350

Common difference = d = 9

Using formula a= a + (n−1) d, to find 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 formula, Sn=n/2(2a+ (n−1) d) to find sum of n terms of AP and putting 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.

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

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

It means a22=149

Using formula an=a+ (n−1) d,   to find nth term of AP, we can say that

149=a+ (22−1) (7)

⇒149=a+147

⇒a=2

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

S22= 22/2(4+ (22−1) (7))

⇒S22=11(4+147)

⇒S22=1661

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

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

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

Using formula an=a+ (n−1) d,   to find nth 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 value of a in equation (2), we can say that

18=14−d+2d

⇒d=4

Therefore, common difference d=4

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

18=a+2(4)

⇒a=10

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

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

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

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

Solution: Let the numbers are 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 numbers are 2,4,6,and 8

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

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

tn = t+ (n-1) d ⇒100 + (n-1)(-3) = 100 – 3(n-1) = 103 – 3n.

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

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

That is, the 35th term will be negative.

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

= 17(100+100+ (34−1) (−3)) =1717

1. 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

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 common difference being 3.

Sum of an AP =  n/2(a+l)

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

In this case, therefore,

998 = 101 + (n – 1) × 3

⇒897= (n−1) ×3

⇒299= (n−1)

⇒n=300

Sum of the AP will therefore, be (101+198)/ 2 ×300= 164, 850

## CBSE Class 10 Maths Chapter 5 Extra MCQs

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

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

In this chapter, concepts discussed include pattern in succeeding term obtained by adding a fixed number to the preceding terms. Students will also see how to find nth terms and the sum of n consecutive terms. Students 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 you to score well in the Class 10 exams.