# 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)Â 7,13,19….,205(ii)Â 18,31/2,13,…âˆ’47
1. 26, 35
2. 27, 34
3. 35, 26
4. 34, 27

Solution: (i) 7, 13, 19…., 205

First term,Â a=7

Common difference,Â d=13âˆ’7=6

an=205

Using formulaÂ an=a+ (nâˆ’1) dÂ to findÂ nthÂ term of arithmetic progression,Â we get

205=7+ (nâˆ’1)6

â‡’205=6n+1

â‡’204=6n

â‡’n=204/6=34

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

(ii)Â 18,31/2,13,…âˆ’47

First term,Â a=18

Common difference,Â d= (31/2) âˆ’18=âˆ’5/2

anÂ = âˆ’ 47

Using formulaÂ an=a+ (nâˆ’1) dÂ to findÂ nthÂ term of arithmetic progression, we get

âˆ’47=18 + (nâˆ’1) (âˆ’5/2)

â‡’âˆ’94=36âˆ’5n+5

â‡’5n=135

â‡’n=135/5=27

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

1. Find theÂ 31stÂ term of an AP whoseÂ 11thÂ term is 38 andÂ 16thÂ term is 73.
1. 185
2. 210
3. 178
4. 150

Solution: We are given that a11=38Â andÂ a16=73Â where, a11Â is theÂ 11thÂ term andÂ Â a16Â is theÂ 16thÂ term of an AP.

Using formulaÂ an=a+ (nâˆ’1) dÂ to findÂ nthÂ term of arithmetic progression, we get

38=a+10d….. (i)

73=a+15dÂ …… (ii)

equation (ii) – equation (i) gives,

35 = 5d

d = 7….. (iii)

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

a31=âˆ’32+(31âˆ’1)(7)

â‡’âˆ’32+210=178

Therefore,Â 31stÂ 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 thatÂ 3rdÂ andÂ 9thÂ term of AP are 4 and -8 respectively.

It meansÂ a3=4Â andÂ a9=âˆ’8

Where,Â a3Â andÂ a9Â are third and ninth terms respectively.

Using formulaÂ an=a+(nâˆ’1)dÂ to findÂ nthÂ term of arithmetic progression, we get

4=a+(3âˆ’1)d

â‡’4Â =Â a+2d…(i)

âˆ’8Â =Â a+(9âˆ’1)d

â‡’âˆ’8=a+8d…(ii)

From equation (i)Â Â we haveÂ Â a=4âˆ’2d

Substituting in equation (ii) , we have

âˆ’8=4âˆ’2d+8d

â‡’âˆ’12Â =Â 6d

â‡’dÂ =âˆ’12/6Â =âˆ’2

Solving for (a) , we get

â‡’âˆ’8Â =Â aâˆ’16

â‡’aÂ =Â 8

Therefore, first termÂ a = 8Â and Common DifferenceÂ d =âˆ’2

We knowÂ Â anÂ =Â a+ (nâˆ’1) dÂ (whereÂ anÂ is theÂ nthÂ term)

Finding value of n whereÂ an=0

0=8+ (nâˆ’1) (âˆ’2)

â‡’0Â =Â 8âˆ’2n+2

â‡’0Â =Â 10âˆ’2n

â‡’2nÂ =Â 10

â‡’nÂ =Â 10/2=5

Therefore,Â 5thÂ term is equal toÂ 0.

1. Which term of the AP: 3, 15, 27, 39, â€¦ will be 132 more than itsÂ 54thÂ term?
1. 70
2. 65
3. 80
4. 55

Solution: Letâ€™s first calculateÂ 54thÂ of the given AP.

First term = a = 3

Common difference = d =15 – 3 = 12

Using formulaÂ an=a+ (nâˆ’1) d, Â Â to findÂ nthÂ Â term of arithmetic progression,Â we get

a54=a+ (54âˆ’1) d

a54Â = 3 + 53 (12) =3 + 636=639

We want to find which term is 132 more than itsÂ 54thÂ term. Letâ€™s suppose it isÂ nthÂ term which is 132 more thanÂ 54thÂ term.

Therefore, we can say that

anÂ =Â a54+132

an=a+ (nâˆ’1) d=3+ (nâˆ’1) (12)}

â‡’3+ (nâˆ’1)12=639+132

â‡’3+12nâˆ’12=771

â‡’12nâˆ’9=771

â‡’12n=780

â‡’n=Â 780/12=65

Therefore,Â 65thÂ term is 132 more than itsÂ 54thÂ term.

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

Solution: Let first term of first AP =Â a

Let first term of 2nd AP =Â aâ€²

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

It is given that difference between theirÂ 100thÂ terms is 100.Â Using formulaÂ anÂ = a + (nâˆ’1) d, Â to findÂ nthÂ term ofÂ arithmetic progression, we can say that

a+ (100âˆ’1) dâˆ’ (aâ€² +(100âˆ’1)d)=a+99dâˆ’aâ€²âˆ’99d=100

â‡’aâˆ’aâ€²=100Â  Â  Â  ……..Â (1)

We want to find difference between theirÂ 1000thÂ terms which means we want to calculate:

a+ (1000âˆ’1) dâˆ’(aâ€²+(1000âˆ’1)d)=a+999dâˆ’aâ€²âˆ’999d=aâˆ’aâ€²

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

a+ (1000âˆ’1) dâˆ’(aâ€²+(1000âˆ’1)d)=a+999dâˆ’aâ€²-999d=aâˆ’aâ€²=100

Therefore, difference between theirÂ 1000thÂ 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 APÂ starting atÂ 105Â because it is the first three digit number divisible byÂ 7.

AP will end atÂ 994Â because it is the last three digit number divisible byÂ 7.

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

First term, a = 105

Common difference,Â dÂ = 112 – 105 = 7

Using formulaÂ an=a+ (nâˆ’1) d, to findÂ Â nthÂ term of arithmetic progression,Â we can say that

994=105+ (nâˆ’1) (7)

â‡’994=105+ (nâˆ’1) (7)

â‡’889=7(nâˆ’1)

â‡’nâˆ’1=889/7â‡’n=127+1=128

994Â is theÂ 128thÂ term of AP. Therefore, there areÂ 128Â 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 theÂ nthÂ terms of two APâ€™s: 63, 65, 67 …. and 3, 10, 17,.. equal?
1. 11
2. 14
3. 12
4. 13

Solution: Letâ€™s first consider AP:Â 63, 65, 67…..

First termÂ =a=63

Common differenceÂ =d=65âˆ’63=2

Using formulaÂ an=a+ (nâˆ’1) d, Â  to findÂ Â nthÂ term of arithmetic progression, we can say that

an=63+ (nâˆ’1) (2)Â Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â (1)

Now, consider second APÂ 3, 10, 17…

First term = a = 3

Common difference = d = 10 -3 = 7

Using formulaÂ an=a+ (nâˆ’1) d, Â  to findÂ Â nthÂ term of arithmetic progression, we can say that

an=3+(nâˆ’1)(7)Â Â  Â  Â  Â  Â  Â  Â  Â  Â  Â (2)

According to the given condition, we can write

(1) = (2)

â‡’63+ (nâˆ’1) (2) =3+ (nâˆ’1) (7)

â‡’63+2nâˆ’2=3+7nâˆ’7

â‡’65=5n

â‡’n=65/5=13

Therefore,Â 13thÂ terms of both the AP’s are equal.

### Introduction to AP

1. IfÂ (x+1),3xÂ andÂ (4x+2)Â are first three terms of an AP, then itsÂ 5thÂ term is :
1. 14
2. 19
3. 24
4. 28

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

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

Hence,Â 3xâ€“(x+1) = (4x+2)â€“3x

â‡’2xâˆ’1=x+2

â‡’x=3

So, the first term,

a=(x+1) =4.

TheÂ common difference,

d=9â€“4=5.

TheÂ nthÂ term of an AP is given by

tn =a+ (nâˆ’1) d

â‡’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. Â½
2. Â¼
3. 4
4. 1

Solution: LetÂ S10Â be the sum of first 10 terms andÂ S5Â be theÂ sum of first 5 terms.

We know,

Sn = (n/2) [2a + (n-1) d]

Given,

S10=4S5

â‡’ (10/2) [2a + (10-1) d] = 4 Ã— (5/2) [2a + (5-1) d]

â‡’ (10/2) [2a+ 9d] = 4 Ã— (5/2) [2a+ 4d]

â‡’ 2a+ 9d = 4a+ 8d

â‡’ a/d= Â½

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

(i) Â  Â  20,Â 40, 60, 80, 100,..

(ii) Â  Â 5, 0, -5, -10, -15, …

(iii) Â  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) Â 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, 16â€¦â€¦..

(ii) 2, 3, 5, 7, 11â€¦â€¦.

(iii) -1, -1.25, -1.5, -1.75â€¦â€¦

(iv) 1, -1, -3, -5, -7â€¦â€¦â€¦â€¦

Answer: (iii) -1, -1.25, -1.5, -1.75â€¦â€¦ and

(iv) 1, -1, -3, -5, -7â€¦â€¦â€¦â€¦

Solution: Consider each list of numbers:

(i) 2, 4, 8, 16â€¦â€¦..

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

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

SinceÂ a2â€“a1â‰ a3â€“a2, this sequence is not an AP

(ii) 2, 3, 5, 7, 11â€¦â€¦..

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

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

SinceÂ a2â€“a1â‰ a3â€“a2, this sequence is not an AP

(iii) -1, -1.25, -1.5, -1.75â€¦â€¦.

Difference between the first two terms = -1.25 â€“ (-1) = -0.25

Difference between the third and second termÂ = -1.5 â€“ (-1.25) = -0.25

Difference between the fourth and third termÂ = -1.75 â€“ (-1.5) = -0.25

SinceÂ a2â€“a1=a3âˆ’a2=a4âˆ’a3, this sequence is an AP

(iv) 1, -1, -3, -5, -7â€¦â€¦â€¦â€¦

Difference between the first two terms = -1 â€“ 1 = -2

Difference between the third and second termÂ = -3 â€“ (-1) = -2

Difference between the fourth and third termÂ = -5 â€“ (-3) = -2

SinceÂ a2â€“a1=a3âˆ’a2=a4âˆ’a3, this sequence is an AP

1. What are the conditions for a sequenceÂ 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,Â a2,Â a3,Â a4,Â a5,Â a6,Â a7,Â a8â€¦ be a sequence.

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

a2â€“a1=a3â€“a2=a4â€“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 wellÂ a, a+d, a+2d…

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

(i) 4, 4 +Â âˆš3Â , 4 + 2âˆš3, 4 + 3âˆš3, 4 + 4âˆš3

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

(iii)Â 3/5, 6/5,Â 9/5,Â 12/5, 3

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

Solution: Consider each series (i) 4, 4 +Â âˆš3, 4 + 2Â âˆš3, 4 + 3Â âˆš3, 4 + 4Â âˆš3

Difference between first two consecutive terms = 4 +Â âˆš3Â – 4 =Â âˆš3

Difference between third and second consecutive terms = 4 + 2Â âˆš3Â – 4 +Â âˆš3Â =Â Â âˆš3

Difference between fourth and third consecutive terms = 4 + 3Â âˆš3â€“ 4 + 2Â âˆš3Â =Â Â âˆš3

SinceÂ a2â€“a1Â =Â a3âˆ’a2Â =Â a4âˆ’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

SinceÂ Â a2â€“a1Â â‰ Â a3âˆ’a2

This series is not an AP

(iii)Â 3/5,Â 6/5,Â 9/5,Â 12/5, 3

Difference between first two consecutive terms =Â (6/5)â€“(3/5)Â =Â 3/5

Difference between third and second consecutive terms =Â (9/5)â€“(6/5)Â =Â 3/5

Difference between fourth and third consecutive terms =Â (12/5)Â â€“Â (9/5)Â =Â 3/5

Since Â a2â€“a1Â =Â a3â€“a2Â =Â a4â€“a3

This series is an AP

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

Difference between first two consecutive terms = -1/5Â â€“ (-Â 1/5) = 0

Difference between third and second consecutive terms = -1/5Â â€“ (-1/5) = 0

Difference between fourth and third consecutive terms = -1/5Â â€“ (-1/5) = 0

SinceÂ a2â€“a1Â =Â a3â€“a2Â =Â a4â€“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,Â a=5

Last term,Â l=50

Sn=440

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

440= n/2 (5+50)

â‡’440/55=n/2

â‡’8=n/ 2

â‡’n=16

Applying formula,Â Sn=n/2(2a+ (nâˆ’1) d)Â to findÂ sum ofÂ nÂ terms of APÂ and putting value ofÂ n,Â we get

440=16/2(2(5) + (16âˆ’1) d)

â‡’440=8(10+15d)

â‡’10+15d=55

â‡’15d=45

â‡’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 =Â a = 17

Last term = lÂ = 350

Common difference =Â d = 9

Using formulaÂ anÂ = a + (nâˆ’1) d,Â to findÂ nth term of arithmetic progression,Â we can say that

350=17+ (nâˆ’1) (9)

â‡’350=17+9nâˆ’9

â‡’342=9n

â‡’n=342/9=38

Applying formula,Â Sn=n/2(2a+ (nâˆ’1) d)Â to findÂ sum of n terms of APÂ and putting value of n, we get

S38=38/2(34+ (38âˆ’1) (9))

â‡’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 andÂ theÂ 22ndÂ term is 149.
1. 1623
2. 1712
3. 1542
4. 1661

Solution: It is given thatÂ 22ndÂ term is equal to 149.

It meansÂ a22=149

Using formulaÂ an=a+ (nâˆ’1) d, Â  to findÂ nthÂ term of AP,Â we can say that

149=a+ (22âˆ’1) (7)

â‡’149=a+147

â‡’a=2

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

S22=Â 22/2(4+ (22âˆ’1) (7))

â‡’S22=11(4+147)

â‡’S22=1661

Therefore, sum of first 22 terms of AP is equal toÂ 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 formulaÂ an=a+ (nâˆ’1) d, Â  to findÂ nth term of arithmetic progression,Â we can say that

14=a+ (2âˆ’1) d

â‡’14=a+ dÂ Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â Â (1)

And, Â  Â 18=a+ (3âˆ’1) d

â‡’18=a+2dÂ Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â Â (2)

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

Using equationÂ (1),Â we can say thatÂ a=14âˆ’d

Putting value ofÂ aÂ in equationÂ (2),Â we can say that

18=14âˆ’d+2d

â‡’d=4

Therefore, common differenceÂ d=4

Putting value ofÂ dÂ in equation numberÂ (1), we can say that

18=a+2(4)

â‡’a=10

Applying formula,Â Sn= n/2(2a+ (nâˆ’1) d)Â to findÂ sum of n terms of AP,Â we get

S15=51/2(20+ (51âˆ’1) 4) =51/2(20+200) =51Ã—110 =5610

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

1. The sum of fourÂ consecutive numbers inÂ 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 areÂ aâˆ’3d,aâˆ’d,a+d,a+3d

Given:Â aâˆ’3d+aâ€”d+a+d+a+3d=20

4a=20

a=5Â and

(aâˆ’3d)2+Â (aâˆ’d)2Â +Â (a+3d)2Â +Â (a+3d)2=120

(4)(a)2Â + 20Â d2=120

(4)(5)2+20Â d2=120

d2=1

d=+1Â orâˆ’1

HenceÂ numbersÂ areÂ 2,4,6,andÂ 8

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

Solution: Here,Â t1Â = 100, common difference, d =Â t2Â â€“Â t1Â = 97 â€“ 100 = -3

tnÂ =Â t1Â + (n-1) d â‡’100 + (n-1)(-3) = 100 â€“ 3(n-1) = 103 â€“ 3n.

LetÂ tnÂ be the first negative term. i.e,

tnÂ < 0Â â‡’Â 103 – 3n < 0Â â‡’Â n >Â 103/3Â > 34

That is, theÂ 35thÂ term will be negative.

Sum of the firstÂ 34Â termsÂ = (34/2) (first term + last term)

=Â 17(100+100+ (34âˆ’1) (âˆ’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 =Â Â n/2(a+l)

we know that in an A.P., the nth termÂ an=a1+ (nâˆ’1)Ã—d

In this case, therefore,

998 = 101 + (n – 1)Â Ã—Â 3

â‡’897= (nâˆ’1) Ã—3

â‡’299= (nâˆ’1)

â‡’n=300

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

## CBSE Class 10 Maths Chapter 5 Extra MCQs

1.Â In an AP, if d = â€“4, n = 7, anÂ = 4. Calculate a.
(a) 20
(b) 7
(c) 28
(d) 6

2.WhatÂ 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 BYJUâ€™S to get complete assistance for CBSE class 10 board exams. At BYJUâ€™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.