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# NCERT Solution for Class Xth Maths Chapter 5 Arithmetic Progressions

Arithmetic Progressions, the fifth chapter of the NCERT Class Xth Maths textbook, is an important topic in the study of mathematics. This chapter deals with the concept of arithmetic progression and its various applications in solving mathematical problems. An arithmetic progression is a sequence of numbers in which each term is obtained by adding a fixed value to the previous term. In this chapter, students will learn about the basic concepts of arithmetic progression, including the nth term of an arithmetic progression, the sum of the first n terms of an arithmetic progression, and the application of arithmetic progression in solving real-world problems.

## Answers of Maths NCERT solutions for Class Xth Maths Chapter 5 Arithmetic Progressions

Unit 5

Arithmetic Progression

Exercise 5.1

Question 1:

In which of the following situations, does the list of numbers involved make as arithmetic progression and why?

1. The taxi fare after each km when the fare is Rs 15 for the ﬁrst km and Rs 8 for each additional km.
2. The amount of air present in a cylinder when a vacuum pump removes of the air remaining in the cylinder at a time.
3. The cost of digging a well after every metre of digging, when it costs Rs 150 for the ﬁrst metre and rises by Rs 50 for each subsequent metre.
4. The amount of money in the account every year, when Rs 10000 is deposited at compound interest at 8% per annum.

1. It can be observed that Taxi fare for 1st km = 15

Taxi fare for ﬁrst 2 km = 15 + 8 = 23

Taxi fare for ﬁrst 3 km = 23 + 8 = 31

Taxi fare for ﬁrst 4 km = 31 + 8 = 39

Clearly 15, 23, 31, 39 … forms an A.P. because every term is 8 more than the       preceding term.

1. Let the initial volume of air in a cylinder be V lit. In each stroke, the vacuum pump removes of air remaining in the cylinder at a time. In other words, after every stroke, only part of air will remain.

Therefore, volumes  will be

Clearly, it can be observed that the adjacent terms of this series do not have the same difference between them. Therefore, this is not an A.P.

1. Cost of digging for ﬁrst metre = 150

Cost of digging for ﬁrst 2 metres = 150 + 50 = 200

Cost of digging for ﬁrst 3 metres = 200 + 50 = 250

Cost of digging for ﬁrst 4 metres = 250 + 50 = 300

Clearly, 150, 200, 250, 300 … forms an A.P. because every term is 50 more than the preceding term.

1. We know that if Rs P is deposited at r% compound interest per annum for n years, our money will be after n years.

Therefore, after every year, our money will be

Clearly, adjacent terms of this series do not have the same difference between them. Therefore, this is not an A.P.

Question 2:

Write ﬁrst four terms of the A.P. when the ﬁrst term a and the common difference d are given as follows

(i) a = 10, d = 10

1. a = − 2, d = 0
2. a = 4, d = − 3
3. a = − 1 d =

(v) a = − 1.25, d = − 0.25

(i) a = 10, d = 10

Let the series be a1, a2, a3, a4, a5 … a1 = a = 10

a2 = a1 + d = 10 + 10 = 20 a3 = a2 + d = 20 + 10 = 30 a4 = a3 + d = 30 + 10 = 40

a5 = a4 + d = 40 + 10 = 50

Therefore, the series will be 10, 20, 30, 40, 50 …

First four terms of this A.P. will be 10, 20, 30, and 40.

1. a = −2, d = 0

Let the series be a1, a2, a3, a4 … a1 = a = −2

a2 = a1 + d = − 2 + 0 = −2 a3 = a2 + d = − 2 + 0 = −2

a4 = a3 + d = − 2 + 0 = −2

Therefore, the series will be −2, −2, −2, −2 …

First four terms of this A.P. will be −2, −2, −2 and −2.

1. a = 4, d = −3

Let the series be a1, a2, a3, a4 … a1 = a = 4

a2 = a1 + d = 4 − 3 = 1 a3 = a2 + d = 1 − 3 = −2

a4 = a3 + d = − 2 − 3 = −5

Therefore, the series will be 4, 1, −2 −5 …

First four terms of this A.P. will be 4, 1, −2 and −5.

1. a = −1, d =

Let the series be a1, a2, a3, a4

Clearly, the series will be

………….

First four terms of this A.P. will be and .

(v) a = −1.25, d = −0.25

Let the series be a1, a2, a3, a4

a1 = a = −1.25

a2 = a1 + d = − 1.25 − 0.25 = −1.50

a3 = a2 + d = − 1.50 − 0.25 = −1.75

a4 = a3 + d = − 1.75 − 0.25 = −2.00

Clearly, the series will be 1.25, −1.50, −1.75, −2.00 ……..

First four terms of this A.P. will be −1.25, −1.50, −1.75 and −2.00.

Question 3:

For the following A.P.s, write the ﬁrst term and the common difference. (i) 3, 1, − 1, − 3 …

(ii) − 5, − 1, 3, 7 …

(iii)

(iv) 0.6, 1.7, 2.8, 3.9 …

(i) 3, 1, −1, −3 …

Here, ﬁrst term, a = 3

Common difference, d = Second term − First term

= 1 − 3 = −2

(ii) −5, −1, 3, 7 …

Here, ﬁrst term, a = −5

Common difference, d = Second term − First term

= (−1) − (−5) = − 1 + 5 = 4

(iii)

Here, ﬁrst term,

Common difference, d = Second term − First term

(iv) 0.6, 1.7, 2.8, 3.9 …

Here, ﬁrst term, a = 0.6

Common difference, d = Second term − First term

= 1.7 − 0.6

= 1.1

Question 4:

Which of the following are APs? If they form an A.P. ﬁnd the common difference d and write three more terms.

(i) 2, 4, 8, 16 …

(ii)

(iii) −1.2, − 3.2, − 5.2, − 7.2 …

(iv) −10, − 6, − 2, 2 …

1. (vi) 0.2, 0.22, 0.222, 0.2222 ….

(vii) 0, − 4, − 8, − 12 …

(viii)      (ix)     1, 3, 9, 27 …

(x) a, 2a, 3a, 4a …

(xi) a, a2, a3, a4 … (xii)

(xiii)

(xiv) 12, 32, 52, 72

(xv) 12, 52, 72, 73 …

(i) 2, 4, 8, 16 …

It can be observed that a2 − a1 = 4 − 2 = 2

a3 − a2 = 8 − 4 = 4

a4 − a3 = 16 − 8 = 8

i.e., ak+1− ak is not the same every time. Therefore, the given numbers are not forming an A.P.

(ii)

It can be observed that

i.e., ak+1− ak is same every time.

Therefore, and the given numbers are in A.P.

Three more terms are

(iii) −1.2, −3.2, −5.2, −7.2 …

It can be observed that

a2 − a1 = (−3.2) − (−1.2) = −2

a3 − a2 = (−5.2) − (−3.2) = −2

a4 − a3 = (−7.2) − (−5.2) = −2

i.e., ak+1− akis same every time. Therefore, d = −2 The given numbers are in A.P.

Three more terms are a5 = − 7.2 − 2 = −9.2

a6 = − 9.2 − 2 = −11.2

a7 = − 11.2 − 2 = −13.2

(iv) −10, −6, −2, 2 …

It can be observed that a2 − a1 = (−6) − (−10) = 4

a3 − a2 = (−2) − (−6) = 4

a4 − a3 = (2) − (−2) = 4

i.e., ak+1 − akis same every time. Therefore, d = 4 The given numbers are in A.P.

Three more terms are a5 = 2 + 4 = 6

a6 = 6 + 4 = 10

a7 = 10 + 4 = 14

(v)

It can be observed that

i.e., ak+1 − ak is same every time. Therefore,

The given numbers are in A.P. Three more terms are

(vi) 0.2, 0.22, 0.222, 0.2222 ….

It can be observed that

a2 − a1 = 0.22 − 0.2 = 0.02

a3 − a2 = 0.222 − 0.22 = 0.002

a4 − a3 = 0.2222 − 0.222 = 0.0002

i.e., ak+1 − ak is not the same every time. Therefore, the given numbers are not in A.P.

(vii) 0, −4, −8, −12 …

It can be observed that a2 − a1 = (−4) − 0 = −4

a3 − a2 = (−8) − (−4) = −4

a4 − a3 = (−12) − (−8) = −4

i.e., ak+1 − ak is same every time. Therefore, d = −4 The given numbers are in A.P.

Three more terms are a5 = − 12 − 4 = −16

a6 = − 16 − 4 = −20

a7 = − 20 − 4 = −24

(viii)

It can be observed that

i.e., ak+1 − ak is same every time. Therefore, d = 0 The given numbers are in A.P.

Three more terms are

(ix) 1, 3, 9, 27 …

It can be observed that a2 − a1 = 3 − 1 = 2

a3 − a2 = 9 − 3 = 6

a4 − a3 = 27 − 9 = 18

i.e., ak+1 − ak is not the same every time. Therefore, the given numbers are not in A.P.

(x) a, 2a, 3a, 4a …

It can be observed that

a2 − a1 = 2a − a = a a3 − a2 = 3a − 2a = a a4 − a3 = 4a − 3a = a

i.e., ak+1 − ak is same every time. Therefore, d = a The given numbers are in A.P.

Three more terms are a5 = 4a + a = 5a

a6 = 5a + a = 6a

a7 = 6a + a = 7a

(xi) a, a2, a3, a4

It can be observed that a2 − a1 = a2 − a = a (a − 1)

a3 − a2 = a3 − a2 = a2 (a − 1)

a4 − a3 = a4 − a3 = a3 (a − 1)

i.e., ak+1 − ak is not the same every time. Therefore, the given numbers are not in A.P.

(xii)

It can be observed that

i.e., ak+1 − ak is same every time. Therefore, the given numbers are in A.P.

And,

Three more terms are

(xiii)

It can be observed that

i.e., ak+1 − akis not the same every time. Therefore, the given numbers are not in A.P.

(xiv) 12, 32, 52, 72

Or, 1, 9, 25, 49 …..

It can be observed that a2 − a1 = 9 − 1 = 8

a3 − a2 = 25 − 9 = 16

a4 − a3 = 49 − 25 = 24

i.e., ak+1 − ak is not the same every time. Therefore, the given numbers are not in A.P.

(xv) 12, 52, 72, 73 …

Or 1, 25, 49, 73 …

It can be observed that a2 − a1 = 25 − 1 = 24

a3 − a2 = 49 − 25 = 24

a4 − a3 = 73 − 49 = 24

i.e., ak+1 − ak is same every time. Therefore, the given numbers are in A.P. And, d = 24

Three more terms are a5 = 73+ 24 = 97

a6 = 97 + 24 = 121

a7 = 121 + 24 = 145

Exercise 5.2

Question 1:

Fill in the blanks in the following table, given that a is the ﬁrst term, d the common difference and an the nth term of the A.P.

I. a = 7, d = 3, n = 8, an = ?

We know that,

For an A.P.

= 7 + (8 − 1) 3

= 7 + (7) 3

= 7 + 21 = 28

Hence, an = 28

1. Given that

a = −18, n = 10, an = 0, d = ?

We know that,

18 = 9d

Hence, common difference, d = 2

1. Given that

d = −3, n = 18, an = −5

We know that,

Hence, a = 46

IV. a = −18.9, d = 2.5, an = 3.6, n = ?

We know that,

Hence, n = 10

V. a = 3.5, d = 0, n = 105, an = ?

We know that, an = a + (n − 1) d

Hence, an = 3.5

Question 2:

Choose the correct choice in the following and justify

I. 30th term of the A.P: 10, 7, 4,…, is

A. 97 B. 77 C. − 77 D. − 87

II 11th term of the A.P. is

A. 28 B. 22 C. − 38 D.

I. Given that

A.P. 10, 7, 4, …

First term, a = 10

Common difference, d = a2 − a1 = 7 − 10

= −3

We know that,

a30 = 10 + (29) (−3)

a30 = 10 − 87 = −77

Hence, the correct answer is C.

II. Given that, A.P.

First term a = −3

Common difference, d = a2 − a1

We know that,

Question 3:

In the following APs ﬁnd the missing term in the boxes

I.

II.

III.

IV.

V.

I.

For this A.P., a = 2

a3 = 26

We know that,

d = 12

a2 = 2 + (2 − 1) 12

= 14

Therefore, 14 is the missing term.

II.

For this A.P., a2 = 13 and a4 = 3

We know that, an = a + (n − 1) d

a2 = a + (2 − 1) d 13 = a + d (I)

a4 = a + (4 − 1) d 3 = a + 3d (II)

On subtracting (I) from (II), we obtain

−10 = 2d

d = −5

From equation (I), we obtain

13 = a + (−5)

a = 18

a3 = 18 + (3 − 1) (−5)

= 18 + 2 (−5) = 18 − 10 = 8

Therefore, the missing terms are 18 and 8 respectively.

III.

For this A.P.,

We know that,

Therefore, the missing terms are and 8 respectively.

IV.

For this A.P., a = −4 and

a6 = 6

We know that, an = a + (n − 1) d

10 = 5d

d = 2

a2 = a + d = − 4 + 2 = −2

Therefore, the missing terms are −2, 0, 2, and 4 respectively.

V.

For this A.P., a2 = 38

a6 = −22

We know that

an = a + (n − 1) d

a6 = a + (6 − 1) d

−22 = a + 5d (2)

On subtracting equation (1) from (2), we obtain

− 22 − 38 = 4d

−60 = 4d

d = −15

Therefore, the missing terms are 53, 23, 8, and −7 respectively.

Question 4:

Which term of the A.P. 3, 8, 13, 18, … is 78?

3, 8, 13, 18, …

For this A.P.,

a = 3

d = a2 − a1 = 8 − 3 = 5

Let nth term of this A.P. be 78.

78 = 3 + (n − 1) 5

75 = (n − 1) 5

(n − 1) = 15

n = 16

Hence, 16th term of this A.P. is 78.

Question 5:

Find the number of terms in each of the following A.P.

I. 7, 13, 19, …, 205

II.

I. 7, 13, 19, …, 205

For this A.P., a = 7

d = a2 − a1 = 13 − 7 = 6

Let there are n terms in this A.P.

an = 205

We know that

an = a + (n − 1) d

Therefore,

198 = (n − 1) 6

33 = (n − 1)

n = 34

Therefore, this given series has 34 terms in it.

II.

For this A.P.,

Let there are n terms in this A.P. Therefore, an = −47 and we know that,

Therefore, this given A.P. has 27 terms in it.

Question 6:

Check whether − 150 is a term of the A.P. 11, 8, 5, 2, …

For this A.P., a = 11

d = a2 − a1 = 8 − 11 = −3

Let −150 be the nth term of this A.P. We know that,

Clearly, n is not an integer.

Therefore, −150 is not a term of this A.P.

Question 7:

Find the 31st term of an A.P. whose 11th term is 38 and the 16th term is 73

Given that, a11 = 38

a16 = 73

We know that,

Similarly,

On subtracting (1) from (2), we obtain 35 = 5d

d = 7

From equation (1),

38 − 70 = a

a = −32

a31 = a + (31 − 1) d

= − 32 + 30 (7)

= − 32 + 210

= 178

Hence, 31st term is 178.

Question 8:

An A.P. consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term

Given that, a3 = 12

a50 = 106

We know that,

12 = a + 2d (I)

Similarly, (II)

On subtracting (I) from (II), we obtain 94 = 47d

d = 2

From equation (I), we obtain 12 = a + 2 (2)

a = 12 − 4 = 8

a29 = a + (29 − 1) d a29 = 8 + (28)2

a29 = 8 + 56 = 64

Therefore, 29th term is 64.

Question 9:

If the 3rd and the 9th terms of an A.P. are 4 and − 8 respectively. Which term of this A.P. is zero.

Given that, a3 = 4

a9 = −8

We know that, an = a + (n − 1) d

(I)

-8 = a + 8d (II)

On subtracting equation (I) from (II), we obtain

−12 = 6d

d = −2

From equation (I), we obtain 4 = a + 2 (−2)

4 = a − 4

a = 8

Let nth term of this A.P. be zero. an = a + (n − 1) d

2n = 10

n = 5

Hence, 5th term of this A.P. is 0.

Question 10:

If 17th term of an A.P. exceeds its 10th term by 7. Find the common difference.

We know that,

For an A.P.,

Similarly, a10 = a + 9d It is given that

a17 − a10 = 7

(a + 16d) − (a + 9d) = 7

7d = 7

d = 1

Therefore, the common difference is 1.

Question 11:

Which term of the A.P. 3, 15, 27, 39, … will be 132 more than its 54th term?

Given A.P. is 3, 15, 27, 39, …

a = 3

d = a2 − a1 = 15 − 3 = 12 a54 = a + (54 − 1) d

= 3 + (53) (12)

= 3 + 636 = 639

132 + 639 = 771

We have to ﬁnd the term of this A.P. which is 771.

Let nth term be 771.

an = a + (n − 1) d 771 = 3 + (n − 1) 12

768 = (n − 1) 12

(n − 1) = 64

n = 65

Therefore, 65th term was 132 more than 54th term. Alternatively,

Let nth term be 132 more than 54th term.

Question 12:

Two APs have the same common difference. The difference between their 100th term is 100, what is the difference between their 1000th terms?

Let the ﬁrst term of these A.P.s be a1 and a2 respectively and the common difference of these A.P.s be d.

For ﬁrst A.P.,

For second A.P.,

Given that, difference between 100th term of these A.P.s = 100

Therefore, (a1 + 99d) − (a2 + 99d) = 100 a1 − a2 = 100 (1)

Difference between 1000th terms of these A.P.s

(a1 + 999d) − (a2 + 999d) = a1 − a2

From equation (1),

This difference, a1 − a2 = 100

Hence, the difference between 1000th terms of these A.P. will be 100.

Question 13:

How many three digit numbers are divisible by 7

First three-digit number that is divisible by 7 = 105 Next number = 105 + 7 = 112

Therefore, 105, 112, 119, …

All are three digit numbers which are divisible by 7 and thus, all these are terms of an

A.P. having ﬁrst term as 105 and common difference as 7.

The maximum possible three-digit number is 999. When we divide it by 7, the remainder will be 5. Clearly, 999 − 5 = 994 is the maximum possible three-digit number that is divisible by 7.

The series is as follows. 105, 112, 119, …, 994

Let 994 be the nth term of this A.P. a = 105

d = 7

Therefore, 128 three-digit numbers are divisible by 7.

Question 14:

How many multiples of 4 lie between 10 and 250?

First multiple of 4 that is greater than 10 is 12. Next will be 16. Therefore, 12, 16, 20, 24, …

All these are divisible by 4 and thus, all these are terms of an A.P. with ﬁrst term as 12 and common difference as 4.

When we divide 250 by 4, the remainder will be 2. Therefore, 250 − 2 = 248 is divisible by 4.

The series is as follows. 12, 16, 20, 24, …, 248

Let 248 be the nth term of this A.P.

Therefore, there are 60 multiples of 4 between 10 and 250.

Question 15:

For what value of n, are the nth terms of two APs 63, 65, 67, and 3, 10, 17, … equal

63, 65, 67, …

a = 63

d = a2 − a1 = 65 − 63 = 2

nth term of this A.P. = an = a + (n − 1) d

an = 61 + 2n (1)

3, 10, 17, …

a = 3

d = a2 − a1 = 10 − 3 = 7

nth term of this A.P. = 3 + (n − 1) 7 an = 3 + 7n − 7

an = 7n − 4 (2)

It is given that, nth term of these A.P.s are equal to each other. Equating both these equations, we obtain

61 + 2n = 7n − 4

61 + 4 = 5n

5n = 65

n = 13

Therefore, 13th terms of both these A.P.s are equal to each other.

*Question 16:

Determine the A.P. whose third term is 16 and the 7th term exceeds the 5th term by 12.

=a3 = 16

a + (3 − 1) d = 16

a + 2d = 16 (1) a7 − a5 = 12

d = 6

From equation (1), we obtain a + 2 (6) = 16

a + 12 = 16

a = 4

Therefore, A.P. will be 4, 10, 16, 22, …

*Question 17:

Find the 20th term from the last term of the A.P. 3, 8, 13, …, 253

Given A.P.  is 3, 8, 13, …, 253

Common difference for this A.P. is 5.

Therefore, this A.P. can be written in reverse order as

253, 248, 243, …, 13, 8, 3

For this A.P., a = 253

d = 248 − 253 = −5

n = 20

Therefore, 20th term from the last term is 158.

*Question 18:

The sum of 4th and 8th terms of an A.P. is 24 and the sum of the 6th and 10th terms is

44. Find the ﬁrst three terms of the A.P.

We know that,

Similarly, a8 = a + 7d

a6 = a + 5d

a10 = a + 9d

Given that, a4 + a8 = 24

a + 3d + a + 7d = 24

2a + 10d = 24 a + 5d = 12 (1)

a6 + a10 = 44

On subtracting equation (1) from (2), we obtain 2d = 22 − 12

2d = 10

d = 5

From equation (1), we obtain a + 5d = 12

a + 5 (5) = 12

a + 25 = 12

a = −13

a2 = a + d = − 13 + 5 = −8

a3 = a2 + d = − 8 + 5 = −3

Therefore, the ﬁrst three terms of this A.P. are −13, −8, and −3.

*Question 19:

Subba Rao started 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?

It can be observed that the incomes that Subba Rao obtained in various years are in A.P. as every year, his salary is increased by Rs 200.

Therefore, the salaries of each year after 1995 are 5000, 5200, 5400, …

Here, a = 5000 d = 200

Let after nth year, his salary be Rs 7000.

Therefore,

200(n − 1) = 2000

(n − 1) = 10

n = 11

Therefore, in 11th year, his salary will be Rs 7000.

*Question 20:

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

Given that, a = 5

d = 1.75

an = 20.75

n = ?

n − 1 = 9

n = 10

Hence, n is 10.

Exercise 5.3

Question 1:

Find the sum of the following APs.

(i) 2, 7, 12, …., to 10 terms.

(ii) −37, − 33, − 29, …, to 12 terms

(iii)     0.6, 1.7, 2.8, …., to 100 terms

(iv) ,… , to 11 terms

(i) 2, 7, 12, …, to 10 terms For this A.P.,

a = 2

d = a2 − a1 = 7 − 2 = 5 n = 10

We know that,

(ii) −37, −33, −29 ,…, to 12 terms For this A.P.,

a = −37

d = a2 − a1 = (−33) − (−37)

= − 33 + 37 = 4

n = 12

We know that,

(iii) 0.6, 1.7, 2.8 ,…, to 100 terms For this A.P.,

a = 0.6

d = a2 − a1 = 1.7 − 0.6 = 1.1 n = 100

We know that,

(iv) , to 11 terms

For this A.P.,

n = 11

We know that,

Question 2:

Find the sums given below

(i)

(ii) 34 + 32 + 30 + 10

(iii) − 5 + (− 8) + (− 11) + (− 230)

(i)

For this A.P., a = 7

l= 84

Let 84 be the nth term of this A.P. l = a + (n − 1)d

22 = n − 1

n = 23

We know that,

(ii) 34 + 32 + 30 + 10

For this A.P., a = 34

Let 10 be the nth term of this A.P. l = a + (n − 1) d

n = 13

(iii) (−5) + (−8) + (−11) + (−230)

For this A.P., a = −5

l = −230

Let −230 be the nth term of this A.P.

(n − 1) = 75

n = 76

And,

Question 3:

In an AP

1. Given a = 5, d = 3, an = 50, ﬁnd n and Sn.
2. Given a = 7, a13 = 35, ﬁnd d and S13.
3. Given a12 = 37, d = 3, ﬁnd a and S12.
4. Given a3 = 15, S10 = 125, ﬁnd d and a10.
5. Given d = 5, S9 = 75, ﬁnd a and a9.
6. Given a = 2, d = 8, Sn = 90, ﬁnd n and an.
7. Given a = 8, an = 62, Sn = 210, ﬁnd n and d.
8. Given an = 4, d = 2, Sn = − 14, ﬁnd n and a.
9. Given a = 3, n = 8, S = 192, ﬁnd d.
10. Given l = 28, S = 144 and there are total 9 terms. Find a.

1. Given that, a = 5, d = 3, an = 50

As an = a + (n − 1)d,

∴ 50 = 5 + (n − 1)3

45 = (n − 1)3

15 = n − 1

n = 16

1. Given that, a = 7, a13 = 35 As an = a + (n − 1) d,

∴ a13 = a + (13 − 1) d 35 = 7 + 12 d

35 − 7 = 12d

28 = 12d

1. Given that, a12 = 37, d = 3

As

a12 = a + (12 − 1)3

37 = a + 33

a = 4

1. Given that, a3 = 15, S10 = 125 As an = a + (n − 1)d,

a3 = a + (3 − 1)d 15 = a + 2d (i)

On multiplying equation (1) by 2, we obtain 30 = 2a + 4d (iii)

On subtracting equation (iii) from (ii), we obtain

−5 = 5d

d = −1

From equation (i), 15 = a + 2(−1)

15 = a − 2

a = 17

a10 = a + (10 − 1)d

a10 = 17 + (9) (−1)

a10 = 17 − 9 = 8

1. Given that, d = 5, S9 = 75

,

25 = 3(a + 20)

25 = 3a + 60

3a = 25 − 60

an = a + (n − 1)d

a9 = a + (9 − 1) (5)

1. Given that, a = 2, d = 8, Sn = 90

As ,

90 = n [2 + (n − 1)4]

90 = n [2 + 4n − 4]

90 = n (4n − 2) = 4n2 − 2n

4n2 − 2n − 90 = 0

4n2 − 20n + 18n − 90 = 0

4n (n − 5) + 18 (n − 5) = 0

(n − 5) (4n + 18) = 0

Either n − 5 = 0 or 4n + 18 = 0

n = 5 or

However, n can neither be negative nor fractional. Therefore, n = 5

an = a + (n − 1)d

a5 = 2 + (5 − 1)8

= 2 + (4) (8)

= 2 + 32 = 34

1. Given that, a = 8, an = 62, Sn = 210

n = 6

an = a + (n − 1)d

62 = 8 + (6 − 1)d

62 − 8 = 5d

54 = 5d

1. Given that, an = 4, d = 2, Sn = −14

an = a + (n − 1)d 4 = a + (n − 1)2

4 = a + 2n − 2

a + 2n = 6

a = 6 − 2n (i)

−28 = n (a + 4)

−28 = n (6 − 2n + 4) {From equation (i)}

−28 = n (− 2n + 10)

−28 = − 2n2 + 10n 2n2 − 10n − 28 = 0

n2 − 5n −14 = 0

n2 − 7n + 2n − 14 = 0

n (n − 7) + 2(n − 7) = 0

(n − 7) (n + 2) = 0

Either n − 7 = 0 or n + 2 = 0 n = 7 or n = −2

However, n can neither be negative nor fractional. Therefore, n = 7

From equation (i), we obtain a = 6 − 2n

a = 6 − 2(7)

= 6 − 14

= −8

1. Given that, a = 3, n = 8, S = 192

192 = 4 [6 + 7d]

48 = 6 + 7d

42 = 7d

d = 6

1. Given that, l = 28, S = 144 and there are total of 9 terms.

(16) × (2) = a + 28

32 = a + 28

a = 4

Question 4:

How many terms of the AP. 9, 17, 25 … must be taken to give a sum of 636?

Let there be n terms of this A.P.

For this A.P., a = 9

d = a2 − a1 = 17 − 9 = 8

636 = n [9 + 4n − 4]

636 = n (4n + 5)

4n2 + 5n − 636 = 0

4n2 + 53n − 48n − 636 = 0

n (4n + 53) − 12 (4n + 53) = 0

(4n + 53) (n − 12) = 0

Either 4n + 53 = 0 or n − 12 = 0

or n = 12

n cannot be . As the number of terms can neither be negative nor fractional, therefore, n = 12 only.

Question 5:

The ﬁrst term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.

Given that, a = 5

l = 45

Sn = 400

n = 16

l = a + (n − 1) d

45 = 5 + (16 − 1) d

40 = 15d

Question 6:

The ﬁrst and the last term of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?

Given that, a = 17

l = 350

d = 9

Let there be n terms in the A.P. l = a + (n − 1) d

350 = 17 + (n − 1)9

333 = (n − 1)9

(n − 1) = 37

n = 38

Thus, this A.P. contains 38 terms and the sum of the terms of this A.P. is 6973.

Question 7:

Find the sum of ﬁrst 22 terms of an AP in which d = 7 and 22nd term is 149.

d = 7

a22 = 149, S22 = ?

an = a + (n − 1)d

a22 = a + (22 − 1)d 149 = a + 21 × 7

149 = a + 147

a = 2

Question 8:

Find the sum of ﬁrst 51 terms of an AP whose second and third terms are 14 and 18 respectively.

Given that, a2 = 14

a3 = 18

d = a3 − a2 = 18 − 14 = 4

a2 = a + d

14 = a + 4

a = 10

= 5610

Question 9:

If the sum of ﬁrst 7 terms of an AP is 49 and that of 17 terms is 289, ﬁnd the sum of ﬁrst n terms.

Given that, S7 = 49

S17 = 289

7 = (a + 3d)

a + 3d = 7 (i)

Similarly,

17 = (a + 8d)

a + 8d = 17 (ii)

Subtracting equation (i) from equation (ii), 5d = 10

d = 2

From equation (i), a + 3(2) = 7

a + 6 = 7

a = 1

= n2

Question 10:

Show that a1, a2 … , an , … form an AP where an is deﬁned as below

1. an = 3 + 4n
2. an = 9 − 5n

Also ﬁnd the sum of the ﬁrst 15 terms in each case.

1. an = 3 + 4n a1 = 3 + 4(1) = 7

a2 = 3 + 4(2) = 3 + 8 = 11

a3 = 3 + 4(3) = 3 + 12 = 15

a4 = 3 + 4(4) = 3 + 16 = 19

It can be observed that

a2 − a1 = 11 − 7 = 4

a3 − a2 = 15 − 11 = 4

a4 − a3 = 19 − 15 = 4

i.e., ak + 1 − ak is same every time. Therefore, this is an AP with common difference as 4 and ﬁrst term as 7.

= 15 × 35

= 525

1. an = 9 − 5n

a1 = 9 − 5 × 1 = 9 − 5 = 4

a2 = 9 − 5 × 2 = 9 − 10 = −1

a3 = 9 − 5 × 3 = 9 − 15 = −6

a4 = 9 − 5 × 4 = 9 − 20 = −11

It can be observed that a2 − a1 = − 1 − 4 = −5

a3 − a2 = − 6 − (−1) = −5

a4 − a3 = − 11 − (−6) = −5

i.e., ak + 1 − akis same every time. Therefore, this is an A.P. with common difference as

−5 and ﬁrst term as 4.

= −465

Question 11:

If the sum of the ﬁrst n terms of an AP is 4n − n2, what is the ﬁrst term (that is S1)? What is the sum of ﬁrst two terms? What is the second term? Similarly ﬁnd the 3rd, the10th and the nth terms.

Given that, Sn = 4n − n2

First term, a = S1 = 4(1) − (1)2 = 4 − 1 = 3

Sum of ﬁrst two terms = S2

= 4(2) − (2)2 = 8 − 4 = 4

Second term, a2 = S2 − S1 = 4 − 3 = 1

d = a2 − a = 1 − 3 = −2

an = a + (n − 1)d

= 3 + (n − 1) (−2)

= 3 − 2n + 2

= 5 − 2n

Therefore, a3 = 5 − 2(3) = 5 − 6 = −1

a10 = 5 − 2(10) = 5 − 20 = −15

Hence, the sum of ﬁrst two terms is 4. The second term is 1. 3rd, 10th, and nth terms are

−1, −15, and 5 − 2n respectively.

Question 12:

Find the sum of ﬁrst 40 positive integers divisible by 6.

The positive integers that are divisible by 6 are 6, 12, 18, 24 …

It can be observed that these are making an A.P. whose ﬁrst term is 6 and common difference is 6.

a = 6

d = 6 S40 =?

= 20[12 + (39) (6)]

= 20(12 + 234)

= 20 × 246

= 4920

Question 13:

Find the sum of ﬁrst 15 multiples of 8.

The multiples of 8 are 8, 16, 24, 32…

These are in an A.P., having ﬁrst term as 8 and common difference as 8.

Therefore, a = 8 d = 8

S15 =?

= 960

Question 14:

Find the sum of the odd numbers between 0 and 50.

The odd numbers between 0 and 50 are 1, 3, 5, 7, 9 … 49

Therefore, it can be observed that these odd numbers are in an A.P. a = 1

d = 2

l = 49

l = a + (n − 1) d 49 = 1 + (n − 1)2

48 = 2(n − 1)

n − 1 = 24

n = 25

= 625

Question 15:

A contract on construction job speciﬁes a penalty for delay of completion beyond a certain date as follows: Rs. 200 for the ﬁrst day, Rs. 250 for the second day, Rs. 300 for the third day, etc., the penalty for each succeeding day being Rs. 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days.

It can be observed that these penalties are in an A.P. having ﬁrst term as 200 and common difference as 50.

a = 200

d = 50

Penalty that has to be paid if he has delayed the work by 30 days = S30

= 15 [400 + 1450]

= 15 (1850)

= 27750

Therefore, the contractor has to pay Rs 27750 as penalty.

Question 16:

A sum of Rs 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is Rs 20 less than its preceding prize, ﬁnd the value of each of the prizes.

Let the cost of 1st prize be P. Cost of 2nd prize = P − 20 And cost of 3rd prize = P − 40

It can be observed that the cost of these prizes are in an A.P. having common difference as −20 and ﬁrst term as P.

a = P

d = −20

Given that, S7 = 700

a + 3(−20) = 100

a − 60 = 100

a = 160

Therefore, the value of each of the prizes was Rs 160, Rs 140, Rs 120, Rs 100, Rs 80, Rs 60, and Rs 40.

*Question 17:

In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of class I will plant 1 tree, a section of class II will plant 2 trees and so on till class XII. There are three sections of each class. How many trees will be planted by the students?

It can be observed that the number of trees planted by the students is in an AP. 1, 2, 3, 4, 5… 12

First term, a = 1

Common difference, d = 2 − 1 = 1

= 6 (2 + 11)

= 6 (13)

= 78

Therefore, number of trees planted by 1 section of the classes = 78 Number of trees planted by 3 sections of the classes = 3 × 78 = 234 Therefore, 234 trees will be planted by the students.

*Question 18:

A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A of radii 0.5, 1.0 cm, 1.5 cm, 2.0 cm, ……… as shown in ﬁgure. What is the total length of such a spiral made up of thirteen consecutive semicircles?

Semi-perimeter of circle

Therefore, I1, I2, I3 ,i.e. the lengths of the semi-circles are in an A.P.,

S13 =?

We know that the sum of n terms of an a A.P. is given by

= 143

Therefore, the length of such spiral of thirteen consecutive semi-circles will be 143 cm.

*Question 19:

200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on. In how many rows are the 200 logs placed and how many logs are in the top row?

It can be observed that the numbers of logs in rows are in an A.P. 20, 19, 18…

For this A.P., a = 20

d = a2 − a1 = 19 − 20 = −1

Let a total of 200 logs be placed in n rows. Sn = 200

400 = n (40 − n + 1)

400 = n (41 − n)

400 = 41n − n2

n2 − 16n − 25n + 400 = 0

n (n − 16) −25 (n − 16) = 0

(n − 16) (n − 25) = 0

Either (n − 16) = 0 or n − 25 = 0 n = 16 or n = 25

an = a + (n − 1)d

a16 = 20 + (16 − 1) (−1)

a16 = 20 − 15

a16 = 5

Similarly,

a25 = 20 + (25 − 1) (−1)

a25 = 20 − 24

= −4

Clearly, the number of logs in 16th row is 5. However, the number of logs in 25th row is negative, which is not possible.

Therefore, 200 logs can be placed in 16 rows and the number of logs in the 16th row is 5.

*Question 20:

In a potato race, a bucket is placed at the starting point, which is 5 m from the ﬁrst potato and other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line.

A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?

[Hint: to pick up the ﬁrst potato and the second potato, the total distance (in metres) run by a competitor is 2 × 5 + 2 ×(5 + 3)]

The distances of potatoes are as follows. 5, 8, 11, 14…

It can be observed that these distances are in A.P. a = 5

d = 8 − 5 = 3

= 5[10 + 9 × 3]

= 5(10 + 27) = 5(37)

= 185

As every time she has to run back to the bucket, therefore, the total distance that the competitor has to run will be two times of it.

Therefore, total distance that the competitor will run = 2 × 185

= 370 m

Alternatively,

The distances of potatoes from the bucket are 5, 8, 11, 14…

Distance run by the competitor for collecting these potatoes are two times of the distance at which the potatoes have been kept. Therefore, distances to be run are

10, 16, 22, 28, 34,……….

a = 10

d = 16 − 10 = 6 S10 =?

= 5[20 + 54]

= 5 (74)

= 370

Therefore, the competitor will run a total distance of 370 m.

Exercise 5.4

*Question 1:

Which term of the A.P. 121, 117, 113 … is its ﬁrst negative term?

[Hint: Find n for an < 0]

Given A.P. is 121, 117, 113 …

a = 121

d = 117 − 121 = −4

an = a + (n − 1) d

= 121 + (n − 1) (−4)

= 121 − 4n + 4

= 125 − 4n

We have to ﬁnd the ﬁrst negative term of this A.P.

Therefore, 32nd term will be the ﬁrst negative term of this A.P.

Question 2:

The sum of the third and the seventh terms of an A.P is 6 and their product is 8. Find the sum of ﬁrst sixteen terms of the A.P.

We know that, an = a + (n − 1) d a3 = a + (3 − 1) d

a3 = a + 2d

Similarly, a7 = a + 6d

Given that, a3 + a7 = 6 (a + 2d) + (a + 6d) = 6 2a + 8d = 6

a + 4d = 3

a = 3 − 4d (i)

Also, it is given that (a3) × (a7) = 8 (a + 2d) × (a + 6d) = 8

From equation (i),

From equation (i)

Where d is

When d is

When a is 1 and d is

= 76

When and d is

*Question 3:

A ladder has rungs 25 cm apart. (See ﬁgure). The rungs decrease uniformly in length

from 45 cm at the bottom to 25 cm at the top. If the top and bottom rungs m are apart, what is the length of the wood required for the rungs?

[Hint: number of rungs ]

It is given that the rungs are 25 cm apart and the top and bottom m rungs are apart.

∴ Total number of rungs

Now, as the lengths of the rungs decrease uniformly, they will be in an A.P.

The length of the wood required for the rungs equals the sum of all the terms of this A.P. First term, a = 45

Last term, l = 25 n = 11

Therefore, the length of the wood required for the rungs is 385 cm.

*Question 4:

The houses of a row are number consecutively from 1 to 49. Show that there is a value of x such that the sum of numbers of the houses preceding the house numbered x is equal to the sum of the number of houses following it.

Find this value of x.

[Hint Sx − 1 = S49 − Sx]

The number of houses was 1, 2, 3 … 49

It can be observed that the number of houses are in an A.P. having a as 1 and d also as 1.

Let us assume that the number of xth house was like this. We know that,

Sum of n terms in an A.P.

Sum of number of houses preceding xth house = Sx − 1

Sum of number of houses following xth house = S49 − Sx

It is given that these sums are equal to each other.

However, the house numbers are positive integers. The value of x will be 35 only.

Therefore, house number 35 is such that the sum of the numbers of houses preceding the house numbered 35 is equal to the sum of the numbers of the houses following it.

*Question 5:

A small terrace at a football ground comprises of 15 steps each of which is 50 m

long and built of solid concrete. Each step has a rise of m and a tread of

(See ﬁgure) calculate the total volume of concrete required to build the terrace.

From the ﬁgure, it can be observed that

1st step is m wide,

2nd step is 1 m wide,

3rd step is m wide.

Therefore, the width of each step is increasing by m each time whereas their height 1/4 m and length 50 m remains the same.

Therefore, the widths of these steps are

Volume of concrete in 1st step

Volume of concrete in 2nd step

Volume of concrete in 3rd step

It can be observed that the volumes of concrete in these steps are in an A.P.

Volume of concrete required to build the terrace is 750 m3.

This chapter comes under unit 3 algebra and this unit has 20 marks allotted in the examination. Students can expect an average of 3 questions from arithmetic progressions. Along with Class 10 examinations, this topic is very important from the point of competitive exams.

Sub-topics of Class 10 Chapter 5 Arithmetic Progression
5.1 Introduction In this chapter, we shall discuss patterns which we come across in our day-to-day life in which succeeding terms are obtained by adding a fixed number to the preceding terms. We shall also see how to find their nth terms and the sum of n consecutive terms, and use this knowledge in solving some daily life problems.

5.2 Arithmetic Progressions The topic describes Arithmetic Progressions, its definition and relatable terms along with fine examples. You will also learn about Finite Arithmetic Progressions and Infinite Arithmetic Progressions. The general form of AP is a, a + d, a + 2d, a + 3d,…

5.3 nth Term of an AP
This topic discusses various methods to determine the nth term of an AP. The concepts are explained with different types of problems solving techniques and finding the nth term of an AP. The examples mentioned in the chapter will help you while solving the exercise problems.

5.4 Sum of First n Terms of an AP
The topics discuss different techniques to find the sum of the first n terms of an AP. It also provides suitable examples which show different techniques to find the sum of the first n terms of AP.

5.5 Summary
It gives an overview of the entire chapter and the important topics explained in the entire chapter. By going through the summary part you can cover the entire chapter in a few points which help in memorizing the essential concepts.

List of Exercise from Class 10 Maths Chapter 5 Arithmetic progression
Exercise 5.1– 4 questions 1 MCQ and 3 descriptive type questions
Exercise 5.2– 20 questions, 1 fill in the blanks, 2 MCQ’s, 7 Short answer questions and 10 Long answer questions
Exercise 5.3– 20 Questions 3 fill in the blanks, 4 daily life examples, and 13 descriptive type questions
Exercise 5.4 5 Questions- 5 Long answer questions

NCERT Solution for Class Xth Maths Chapter 5 Arithmetic Progressions is an essential resource for students who are looking to develop a solid understanding of arithmetic progression. Through this chapter, students will gain knowledge about the properties and applications of arithmetic progression, which will help them in solving real-world problems. The NCERT solutions for this chapter are designed to provide students with a comprehensive understanding of the subject matter, with step-by-step solutions for each problem. By using these solutions, students can strengthen their problem-solving skills and improve their overall performance in mathematics. Therefore, we recommend students to make the most of the NCERT solutions for Class Xth Maths Chapter 5 Arithmetic Progressions, and take advantage of the opportunity to develop their mathematical abilities.

In this chapter, students will discuss patterns in succeeding terms obtained by adding a fixed number to the preceding terms. They 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.

1. What topics are covered in NCERT Solution for Class Xth Maths Chapter 5 Arithmetic Progressions? Answer: NCERT Solution for Class Xth Maths Chapter 5 Arithmetic Progressions covers various topics related to arithmetic progressions, including the nth term of an arithmetic progression, the sum of the first n terms of an arithmetic progression, and the application of arithmetic progression in solving real-world problems.
2. How can NCERT Solution for Class Xth Maths Chapter 5 Arithmetic Progressions help students in their preparation for exams? Answer: NCERT Solution for Class Xth Maths Chapter 5 Arithmetic Progressions provides step-by-step solutions for each problem, which helps students in understanding the concepts and strengthens their problem-solving skills. By using these solutions, students can improve their overall performance in mathematics and score better in exams.
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Key Features of NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic progressions
Has answers to different types of questions such as MCQs and long answer questions.
Solving this NCERT solution will make you get well versed with important formulas.
Act as a basis to solve arithmetic progression problems asked in competitive examination.
Has answers to all the exercise questions provided in NCERT textbook
Provide you with the necessary practice of solving questions
You can solve different types of questions with varying difficulty.
Different examples taken from day to day life will help you understand the topic thoroughly.
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What kind of questions are there in NCERT Solutions for Class 10 Maths Chapter 5?
Chapter 5 of NCERT Solutions for Class 10 Maths has multiple choice questions, descriptive type of questions, long answer type questions, short answer type questions, fill in the blanks and daily life examples. By the end of this chapter students can increase their problem solving skills and time management skills. This helps in procuring high marks in their finals.
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Yes, it is compulsory to learn all the topics provided in NCERT Solutions for Class 10 Maths Chapter 5 to score high marks in Class 10 board exams.These solutions are designed by subject matter experts who have assembled model questions covering all the exercise questions from the textbook. They also focus on cracking the Solutions of Maths in such a way that it is easy for the students to understand.
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Yes, it is enough to solve all the questions that come in the board exam of NCERT Solutions for Class 10 Maths Chapter 5. Practising this chapter can make them learn the concepts flawlessly. These questions have been devised, as per the NCERT syllabus and the guidelines. This makes the students to score good marks in the finals.