NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions

NCERT Solutions For Class 10 Maths Chapter 5 Arithmetic Progressions – The team of subject matter experts at GFG have made detailed NCERT Solutions for Class 10 Maths Chapter 6 Triangles to make sure that every student can understand how to solve Arithmetic Progressions problems in a stepwise manner.

Class 10 Chapter 5 Arithmetic Progressions of NCERT Maths textbook covers Arithmetic Progression (AP) principles such as determining the nth term of a series, the sum of the first n terms of a series, and using AP in real-world situations.

Class 10 Maths NCERT Solutions Chapter 5 Arithmetic Progressions Exercises

• NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Exercise 5.3 Set 1Set 2 â€“ 20 Questions (7 Short Answers, 13 Long Answers)

This article provides solutions to all the problems asked in Class 10 Maths Chapter 5 Arithmetic Progressions of your NCERT textbook in a step-by-step manner. They are regularly revised to check errors and updated according to the latest CBSE Syllabus 2023-24 and guidelines.

All of the exercises in Chapter 5 Arithmetic Progressions of your textbook have been properly covered in the NCERT Solutions for Class 10 Maths.

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions: Exercise 5.1

(i). The taxi fare after each km when the fare is Rs. 15 for the first km and Rs. 8 for each additional km.

Solution:

Initially fare = 0

After completing 1 km, fare = 15

After completing 2 km, fare = 15 + 8 = 23

After completing 3 km, fare = 23 + 8 = 31

And so on

So you can write fare in series as

15, 23, 31, 39,……… and so on

Here you can see that the first term is 15 and the difference between any two-term is 8. So, this series is in arithmetic progression.

Note: A series is said to be in arithmetic progression if any term can be found out by adding a fixed number to the previous term. The first term is denoted as and the fixed number is known as a difference which is denoted by d.

In the above series a = 15 and d = 8

(ii). The amount of air present in a cylinder when a vacuum pump removes 1/4 of the air remaining in the cylinder at a time.

Solution:

Initial volume of air in cylinder = Ï€R2H

Amount of air removed by vacuum pump = Ï€R2H/4

Remaining air = 3 * Ï€R2H/4

Again amount of air removed by vacuum pump = 3 * Ï€R2H/16

Remaining air = 3 * Ï€R2H/4 – 3 * Ï€R2H/16 = 9 * Ï€R2H/16

and so on

The series can be written as: Ï€R2H, 3 * Ï€R2H/4, 9*Ï€R2H/16,……… and so on

You can see here is no common difference between any two-term. So, this series is not arithmetic progression.

(iii).  The cost of digging a well after every meter of digging, when it costs Rs. 150 for the first meter and rises by Rs. 50 for each subsequent meter.

Solution:

Cost of digging well for first meter is 150 rupees

Cost of digging well for two meters is 150 + 50 = 200 rupees

Cost of digging well for three meters is 200 + 50 = 250 rupees

And so on

The series for cost of digging a well look like 150, 200, 250, 300,……… and so on

Here you can see that after the first term, any term can be found out by adding 50 to the previous term. Hence, the above series is in arithmetic progression with the first term 150 and the common difference is 50.

(iv). The amount of money in the account every year, when Rs 10000 is deposited at compound interest at 8% per annum.

Solution:

Formula for compound interest is given by P(1 + r/100)n where P is the principal amount, r is the rate of interest and n is the time duration.

So, the series can be written as: 10000(1 + 8/100), 10000(1 + 8/100)2,10000(1 + 8/100)3,……… and so on

As time duration increases, compound interest will increase exponentially. Hence, this can’t form a series in arithmetic series because there is no fixed common difference between any two terms.

(i). a = 10, d = 10

Solution:

Given:

a1 = a = 10

d = 10

Now we find the remaining terms:

Second term(a2) = a1 + d = 10 + 10 = 20

Third term(a3) = a2 + d = 20 + 10 = 30

Fourth term(a4) = a3 + d = 30 + 10 = 40

and so on.

So, four terms of this A.P. will be as follows

10, 20, 30, 40,….

(ii) a = -2, d = 0

Solution:

Given:

a1 = a = -2

d = 0

Now we find the remaining terms:

Second term(a2) = a1 + d = -2 + 0 = -2

Third term(a3) = a2 + d = -2 + 0 = -2

Fourth term(a4) = a3 + d = -2 + 0 = -2

and so on.

So, four terms of this A.P. will be as follows

-2, -2, -2, -2,….

(iii). a = 4, d = -3

Solution:

Given:

a1 = a = 4

d = -3

Now we find the remaining terms:

Second term(a2) = a1 + d = 4 + (-3) = 1

Third term(a3) = a2 + d = 1 + (-3) = -2

Fourth term(a4) = a3 + d = -2 + (-3) = -5

and so on.

So, four terms of this A.P. will be as follows

4, 1, -2, -5,….

(iv). a = -1, d = 1/2

Solution:

Given:

a1 = a = -1

d = 1/2

Now we find the remaining terms:

Second term(a2) = a1 + d = -1 + 1/2 = -1/2

Third term(a3) = a2 + d = (-1/2) + (1/2) = 0

Fourth term(a4) = a3 + d = 0 + (1/2) = 1/2

and so on.

So, four terms of this A.P. will be as follows

-1, -1/2, 0, 1/2,….

(v). a = -1.25, d = -0.25

Solution:

Given:

a1 = a = -1.25

d = -0.25

Now we find the remaining terms:

Second term(a2) = a1 + d = -1.25 – 0.25 = – 1.50

Third term(a3) = a2 + d = -1.50 – 0.25 = -1.75

Fourth term(a4) = a3 + d = -1.75 – 0.25 = -2.00

and so on.

So, four terms of this A.P. will be as follows

-1.25, -1.50, -1.75, -2.00,….

(i). 3, 1, -1, -3,….

Solution:

From the above A.P., first term (a) = 3

Common difference (d) = second term – first term

= 1 – 3 = -2

(ii). -5, -1, 3, 7,….

Solution:

From the above A.P., first term (a) = -5

Common difference (d) = second term – first term

= -1 – (-5)

= -1 + 5 = 4

(iii). 1/3, 5/3, 9/3, 13/3,….

Solution:

From the above A.P., first term (a) = 1

Common difference (d) = second term – first term

= 5/3 – 1/3

= 4/3

(iv). 0.6, 1.7, 2.8, 3.9,….

Solution:

From the above A.P., first term (a) = 0.6

Common difference (d) = second term – first term

= 1.7 – 0.6

= 1.1

(i). 2, 4, 8, 16,….

Solution:

First we check the given series is AP or not by finding common difference:

d1 = a2 + a1 = 4 – 2 = 2

d2 = a3 + a2 = 8 – 4 = 4

So d1 â‰  d2

Hence, this series doesn’t form an AP because there is no fixed common difference.

(ii). 2, 5/2, 3, 7/2,….

Solution:

First we check the given series is AP or not by finding common difference:

d1 = a2 + a1 = 5/2 – 2 = 1/2

d2 = a3 + a2 = 3 – 5/2 = 1/2

So d1 = d2

Hence, the above series is in arithmetic progression and the common difference is 1/2.

Now we find the remaining terms of the A.P.:

Fifth term(a5) = a4 + d = 7/2 + 1/2 = 4

Sixth term(a6) = a5 + d = (4) + (1/2) = 9/2

Seventh term(a7) = a6 + d = 9/2 + (1/2) = 5

So, the next three terms are: 4, 9/2, 5

(iii). -1.2, -3.2, -5.2, -7.2,….

Solution:

First we check the given series is AP or not by finding common difference:

d1 = a2 + a1 = -3.2 – (-1.2) = -2.0

d2 = a3 + a2 = -5.2 – (-3.2) = -2.0

So d1 = d2

Hence, the above series is in arithmetic progression and the common difference is -2.0

Now we find the remaining terms of the A.P.:

Fifth term(a5) = a4 + d = (-7.2) + (-2.0) = -9.2

Sixth term(a6) = a5 + d = (-9.2) + (-2.0) = -11.2

Seventh term(a7) = a6 + d = (-11.2) + (-2.0) = -13.2

So, the next three terms are: -9.2, -11.2, -13.2

(iv). -10, -6, -2, 2,….

Solution:

First we check the given series is AP or not by finding common difference:

d1 = a2 + a1 = -6 – (-10) = 4

d2 = a3 + a2 = -2 – (-6) = 4

So d1 = d2

Hence, the above series is in arithmetic progression and the common difference is 4

Now we find the remaining terms of the A.P.:

Fifth term(a5) = a4 + d = 2 + 4 = 6

Sixth term(a6) = a5 + d = 6 + 4 = 10

Seventh term(a7) = a6 + d = 10 + 4 = 14

So, the next three terms are 6, 10, 14

(v). 3, 3+âˆš2, 3+2âˆš2, 3+3âˆš2,….

Solution:

First we check the given series is AP or not by finding common difference:

d1 = a2 + a1 = 3 + âˆš2 – 3 = âˆš2

d2 = a3 + a2 = 3 + 2âˆš2 – (3 + âˆš2) = âˆš2

So d1 = d2

Hence, the above series is in arithmetic progression and the common difference is âˆš2

Now we find the remaining terms of the A.P.:

Fifth term(a5) = a4 + d = (3 + 3âˆš2) + âˆš2 = 3 + 4âˆš2

Sixth term(a6) = a5 + d = (3 + 4âˆš2) + âˆš2 = 3 + 5âˆš2

Seventh term(a7) = a6 + d = (3 + 5âˆš2) + âˆš2 = 3 + 6âˆš2

So, the next three terms are 3+4âˆš2, 3+5âˆš2, 3+6âˆš2

(vi). 0.2, 0.22, 0.222, 0.2222,….

Solution:

First we check the given series is AP or not by finding common difference:

d1 = a2 + a1 = 0.22 – 0.2 = 0.02

d2 = a3 + a2 = 0.222 – 0.22 = 0.002

So d1 â‰  d2

Hence, this series doesn’t form an AP because there is no fixed common difference.

(vii). 0, -4, -8, -12,….

Solution:

First we check the given series is AP or not by finding common difference:

d1 = a2 + a1 = -4 – 0 = -4

d2 = a3 + a2 = -8 – (-4) = -4

So d1 = d2

Hence, the above series is in arithmetic progression and the common difference is -4

Now we find the remaining terms of the A.P.:

Fifth term(a5) = a4 + d = (-12) + (-4) = -16

Sixth term(a6) = a5 + d = (-16) + (-4) = -20

Seventh term(a7) = a6 + d = (-20) + (-4) = -24

So, the next three terms are -16, -20, -24

(viii). -1/2, -1/2, -1/2, -1/2,….

Solution:

First we check the given series is AP or not by finding common difference:

d1 = a2 + a1 = -1/2 – (-1/2) = 0

d2 = a3 + a2 = -1/2 – (-1/2) = 0

So d1 = d2

Hence, the above series is in arithmetic progression and the common difference is 0

Now we find the remaining terms of the A.P.:

Fifth term(a5) = a4 + d = (-1/2) + 0 = -1/2

Sixth term(a6) = a5 + d = (-1/2) + 0 = -1/2

Seventh term(a7) = a6 + d = (-1/2) + 0 = -1/2

So, the next three terms are -1/2, -1/2, -1/2

(ix). 1, 3, 9, 27,….

Solution:

First we check the given series is AP or not by finding common difference:

d1 = a2 + a1 = 3 – 1 = 2

d2 = a3 + a2 = 9 – 3 = 6

So d1 â‰  d2

Hence, this series doesn’t form an AP because there is no fixed common difference.

(x). a, 2a, 3a, 4a,….

Solution:

First we check the given series is AP or not by finding common difference:

d1 = a2 + a1 = 2a – a = a

d2 = a3 + a2 = 3a – 2a = a

So d1 = d2

Hence, the above series is in arithmetic progression and the common difference is a

Now we find the remaining terms of the A.P.:

Fifth term(a5) = a4 + d = 4a + a = 5a

Sixth term(a6) = a5 + d = 5a + a = 6a

Seventh term(a7) = a6 + d = 6a + a = 7a

So, the next three terms are 5a, 6a, 7a

(xi). a, a2, a3, a4,….

Solution:

First we check the given series is AP or not by finding common difference:

d1 = a2 + a1 = a2 – a

d2 = a3 + a2 = a3 – a2

So d1 â‰  d2

Hence, this series doesn’t form an AP because there is no fixed common difference.

(xii). âˆš2, âˆš8, âˆš18, âˆš32,….

Solution:

First we check the given series is AP or not by finding common difference:

d1 = a2 + a1 = 2âˆš2 – âˆš2 = âˆš2

d2 = a3 + a2 = 3âˆš2 – 2âˆš2 = âˆš2

So d1 = d2

Hence, the above series is in arithmetic progression and the common difference is âˆš2

Now we find the remaining terms of the A.P.:

Fifth term(a5) = a4 + d = âˆš32 + âˆš2 = 5âˆš2

Sixth term(a6) = a5 + d = 5âˆš2 + âˆš2 = 6âˆš2

Seventh term(a7) = a6 + d = 6âˆš2 + âˆš2 = 7âˆš2

So, the next three terms are 5âˆš2, 6âˆš2, 7âˆš2

(xiii). âˆš3, âˆš6, âˆš9, âˆš12,….

Solution:

First we check the given series is AP or not by finding common difference:

d1 = a2 + a1 = âˆš6 – âˆš3 = âˆš3(âˆš2 – 1)

d2 = a3 + a2 = âˆš9 – âˆš6 = âˆš3(âˆš3 – âˆš2)

So d1 â‰  d2

Hence, this series doesn’t form an AP because there is no fixed common difference.

(xiv). 12, 32, 52, 72,….

Solution:

First we check the given series is AP or not by finding common difference:

d1 = a2 + a1 = 9 – 1 = 8

d2 = a3 + a2 = 25 – 9 = 16

So d1 â‰  d2

Hence, this series doesn’t form an AP because there is no fixed common difference.

(xv). 12, 52, 72, 73,….

Solution:

First we check the given series is AP or not by finding common difference:

d1 = a2 + a1 = 25 – 1 = 24

d2 = a3 + a2 = 49 – 25 = 24

So d1 = d2

Hence, the above series is in arithmetic progression and the common difference is 24

Now we find the remaining terms of the A.P.:

Fifth term(a5) = a4 + d = 73 + 24 = 97

Sixth term(a6) = a5 + d = 97 + 24 = 121

Seventh term(a7) = a6 + d = 121 + 24 = 145

So, the next three terms are 97, 121, 145

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions: 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

 a d n an I 7 3 8 – II -18 – 10 0 III – -3 18 -5 IV -18.9 2.5 – 3.6 V 3.5 0 105 –

Solution:

(i) a = 7, d = 3, n = 8

Using A.P formula,

an = a + (n – 1)d

So, an = 7 + (8 – 1) * 3

an = 7 + 21

an = 28

(ii) a = -18, d = ?, n = 10, an = 0

Using A.P formula,

an = a + (n – 1)d

0 = -18 + (10 – 1) * d

18 = 9 * d

d = 2

(iii) a = ?, d = -3, n = 18, an = -5

Using A.P formula,

an = a + (n – 1)d

-5 = a + (18 – 1) * (-3)

-5 = a – 51

a = -5 + 51

a = 46

(iv) a = -18.9, d = 2.5, n = ?, an = 3.6

Using A.P formula,

an = a + (n – 1)d

3.6 = -18.9 + (n – 1) * 2.5

3.6 + 18.9 = (n – 1) * 2.5

n – 1 = 22.5/2.5

n – 1 = 9

n = 10

(v) a = 3.5, d = 0, n = 105, an = ?

Using A.P formula,

an = a + (n – 1)d

an = 3.5 + (105 – 1) * 0

an = 3.5

(A) 97          (B) 77         (C) -77          (D) -87

Solution:

Given: A.P. is 10, 7, 4, ……

Now we find the common difference:

Common difference(d) = second term – first term

So, d = 7 -10 = -3

The first term (a) = 10

Total number of terms (n) = 30

Now, we find the 30th term

a30 (30th term) = a + (n – 1)d

a30 = 10 + (30 – 1) * (-3)

a30 = 10 – 87

a30 = -77

Hence, the correct option is C.

(A) 28          (B) 22          (C) -38          (D)

Solution:

Given: A.P. is -3, -1/2, 2, …….

So, first term (a) = -3

Now, we find the common difference;

Common Difference (d) = second term – first term

d = -1/2 – (-3)

d = -1/2 + 3

d = 5/2

11th term can be calculated by following formula

an = a +(n – 1)d

a11 = -3 + (11 – 1) * (5/2)

= -3 + (10) * (5/2)

= -3 + 25

a11 = 22

Hence, option B is the correct choice.

(i) 2, , 26

Solution :

Given: first term (a) = 2

third term (a3) = 26

a3 can be calculated using the formula an = a + (n – 1)d

a3 = 2 + (3 – 1) * d

26 = 2 + 2d

24 = 2d

d = 12

So a2 can be calculated using the formula an = a + (n – 1)d

a2 = 2 + (2 – 1) * 12

a2 = 2 + 12

a2 = 14

(ii), 13, , 3

Solution:

Given:

a2 = 13

a + (2 – 1)d = 13

a + d = 13         -(1)

a4 = 3

a + (4 – 1)d = 3

a + 3d = 3         -(2)

After solving equation (1) and (2), you will get

d = -5

and a = 18

Now we find a3

a3 = 18 + (3 – 1) * (-5)

a3 = 18 -10

a3 = 8

Hence, the missing terms in the square boxes are 18 and 8.

(iii) 5, , ,

Solution:

Given:

a  = 5

a4 = 19/2

a + (4 – 1)d = 19/2

5 + 3d = 19/2

3d = (19/2) – 5

3d = 9/2

d = 9/6 = 3/2

So, a2 = 5 + (2 – 1) * (3/2)

a2 = 5 + 3/2

a2 = 13/2

and a3 = 5 + (3 – 1) * (3/2)

a3 = 5 + 3

a3 = 8

Hence, the missing terms in the square boxes are 13/2 and 8.

(iv) -4, , , , , 6

Solution:

Given: a = -4

a6 = 6

a + (6 – 1)d = 6

-4 + 5d = 6

5d = 10

d = 2

So, a2 = a + d

a2 = -4 + 2 = -2

a3 = a + 2d

a3 = -4 + 4 = 0

a4 = a + 3d

a4 = -4 + 6 = 2

a5 = a + 4d

a5 = -4 + 8 = 4

Hence, the missing terms in the square boxes are -2, 0, 2, and 4.

(v) , 38, , , , -22

Solution:

Given:

a2 = 38

a + d = 38         -(1)

a6 = -22

a + 5d = -22         -(2)

After solving (1) and (2) equation, you will get

d = -15

and a= 53

So,

a3 = a + 2d

a3 = 53 – 30 = 23

a4 = a+ 3d

a4 = 53 – 45 = 8

a5 = a + 4d

a5 = 53 – 60 = -7

Hence, the missing terms in the square boxes are 53, 23, 8 and -7.

Question 4. Which term of the A.P. 3, 8,13, 18, ……. is 78 ?

Solution:

Given:

a = 3

d = second term – first term

d = 8 – 3 = 5

and an = 78

n = ?

a + (n – 1)d = 78

3 + (n – 1) * 5 = 78

(n – 1) * 5 = 78 – 3

n – 1 = 75 /5

n = 15 +1

n = 16

So, 78 is the 16th term of the given A.P.

(i) 7,13, 19, ……, 205

Solution:

Find: n = ?

Given: a = 7

d = 13 – 7 = 6

an = 205

a + (n – 1)d = 205

7 + (n – 1) * 6 = 205

(n – 1) * 6 = 205 – 7

n – 1 = 198/6

n = 33 + 1

n = 34

So, there are 34 terms in the given A.P.

(ii) 18, , 13, ……., -47

Solution:

Find: n = ?

Given: a = 18

d = 31/2 – 18 = -5/2

an = -47

18 + (n – 1) * (-5/2) = -47

(n – 1) * (-5/2) = -47 – 18

n – 1 = -65 * (-2/5)

n – 1 = 130/5

n = 26 + 1

n = 27

Hence, there are 27 terms in the given A.P.

Question 6. Check whether -150 is a term of the A.P. 11, 8, 5, 2,……..

Solution:

Given:

a = 11

d = 8 – 11 = -3

Suppose -150 is the nth term of the given A.P. then it can be written as

a + (n – 1)d = -150

11 + (n – 1) * (-3) =- 150

(n – 1) * (-3) = -150 – 11

n – 1 = -161/(-3)

n = 161/3 + 1

n = 164/3

As you can see that n is not in integer. Hence, -150 is not a term in the given A.P.

Question 7. Find the 31st term of an A.P. whose 11th term is 38 and the 16th term is 73.

Solution:

Given:

a11 = 38

a + 10 * d = 38         -(1)

a16 = 73

a + 15 * d = 73         -(2)

Subtracting equation (1) from equation (2), you will get

15 * d – 10 * d = 73 – 38

d = 35/5

d = 7

After putting value of d in equation (1) you will the value of a

a = 38 – 70

a = -32

So, a31 = a + (31 – 1) * d

a31 = -32 + 30 * 7

a31 = 178

Hence, 31st term of the given A.P. is 178.

Question 8. In A.P. consists of 50 terms of which the 3rd term is 12 and the last term is 106. Find the 29th term.

Solution

According to the question:

n = 50

a3 = 12

a + 2 * d = 12         -(1)

a50 = 106

a + 49 * d  = 106         -(2)

After solving (1) and (2) you will get

47 * d = 106 – 12

d = 94/47 = 2

and a = 12 -4 = 8

So, a29 = a + 28 * d

a29 = 8 + 28 * 2

a29 = 64

So the 29th term of the given A.P. will be 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?

Solution:

Given: a3 = 4

a + 2 * d = 4         -(1)

and a9 = -8

a + 8 * d = -8          -(2)

After solving (1) and (2), you will get

d = -2

and a = 8

Let the nth term of the A.P. will be zero.

So an = 0

a + (n – 1)d = 0

8 + (n – 1) * (-2) = 0

n – 1 = -8/(-2)

n = 4 + 1

n = 5

Hence, the 5th term of the given A.P. will be zero.

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

Solution:

According to the question

a17 = a10 + 7

a + (17 – 1)d = a + (10 – 1)d + 7

16 * d = 9 * d + 7

7 * d = 7

d = 1

Hence, the common difference of the given A.P. will be 1.

Question 11. Which term of the A.P. 3, 15, 27, 39, ….. will be 132 more than its 54th term?

Solution:

According to the question

d = 15 – 3 = 12

Let the nth term of the A.P. will be 132 more than its 54th term

an = a54 + 132

a + (n – 1)d = a + (54 – 1)d + 132

(n – 1)(12) = 53 * 12 + 132

(n – 1) * 12 = 636 + 132

n – 1 = 771 / 12

n = 64 + 1

n = 65

Hence, 65th term will be 132 more than its 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 term?

Solution:

Let d be common difference of both APs and a1 be the first term of

one A.P. and a2 be the first term of other A.P.

So, a100 for first A.P. will be

a100 = a1 + 99d

and a100 for second A.P. will be

a100 = a2 + 99d

According to the question, the difference between their 100th term is 100

So, (a1 + 99d) – (a2 + 99d) = 100

a1 – a2 = 100         -(1)

Now a100 for first A.P. will be

a1000 = a1 + 999d

and a1000 for second A.P. will be

a1000 = a2 + 999d

Difference between their 1000th term will be

= (a1 + 999d) – (a2 + 999d)

= a1 – a2

= 100             -(from (1) a1 -a2 = 100)

Hence, the difference between their 1000th term will be 100.

Question 13. How many three-digit numbers are divisible by 7?

Solution:

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

So, the first term of the AP (a) = 105

and common difference (d) = 7

The last three-digit number that is divisible by 7 = 994

So AP will look like 105, 112, 119,………, 994

Let there are n three-digit numbers between 105 and 994

You have an = 994

a + (n – 1)d = 994

105 + (n – 1) * 7 = 994

(n – 1) * 7 = 994 – 105

n – 1 = 889/7

n = 127 + 1

n = 128

Hence, there 128 three-digit numbers that are divisible by 7.

Question 14. How many multiples of 4 lie between 10 and 250?

Solution:

Given:

12 is the minimum number that is divisible by 4 between 10 and 250.

So, a = 12

d = 4

248 is the highest number that is divisible by 4 between 10 and 250.

So an = 248

a + (n – 1)d = 248

12 + (n – 1) * 4 = 248

(n – 1) * 4 = 236

n – 1 = 59

n = 60

Hence, 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?

Solution:

For A.P. 63, 65, 67,…..

a = 63 and d = 65 – 63 = 2

nth term for this AP will be

an = 63 + (n – 1) * 2

For A.P. 3, 10, 17, ….

a = 3 and d = 10 – 3 = 7

nth term for this AP will be

an = 3 + (n – 1) * 7

According to the question, nth terms of both APs are equal

So, 63 + (n – 1) * 2 = 3 + (n – 1) * 7

60 = (n – 1) * 5

n – 1 = 12

n =13

Hence, 13th of both the APs are equal.

Question 16. Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.

Solution:

a3 = 16

a + 2d = 16          -(1)

and a7 = a5 + 12

a + 6d = a + 4d + 12

2d = 12

d = 6

On putting value of d in equation (1), you will get

a = 16 – 12

a = 4

So, AP will look like, 4, 10, 16, 22, 28,…..

Question 17. Find the 20th term from the last term of the AP 3, 8, 13,………, 253

Solution:

d = 8 – 3 = 5

In reverse order the AP will be

253, 248, ………. 13, 8, 3

Now for this AP

a = 253 and d = 248 – 253 = -5

So 20th term will be

a20 = a + 19d

a20 = 253 + 19 * (-5)

a20 = 158

Hence, the 20th term from the last term for the given AP will be 158.

Question 18. The sum of 4th and 8th terms of an AP is 24 and the sum of the 6th term and 10th terms is 44. Find the first three terms of the AP.

Solution:

Given: a4 + a8 = 24

a + 3d + a + 7d = 24

2a + 10d = 24

a + 5d = 12         -(1)

and a6 + a10 = 44

a + 5d + a + 9d = 44

2a + 14d = 44

a + 7d = 22         -(2)

From (1) and (2), you will get

d = 5

and a = -13

So a2 = a + d

a2 = -13 + 5 = -8

a3 = a + 2d

a3 = -13 + 10 = -3

Hence, the first three terms of the AP are -13, -8, -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.

Solution:

As, salary of Subba Rao is increasing by a fixed amount in every year hence

this will form an AP with first term (a) = 5000 and common difference (d) d = 200

an = 7000

a +(n – 1)d = 7000

5000 + (n – 1) * 200 = 7000

n – 1 = 2000/200

n = 10 + 1

n = 11

Hence, the salary will be 7000 in 11th year.

Question 20. Ramkali saved Rs 5 in the first week of a year and then increased her weekly saving by 1.75. If in the nth week her weekly savings become Rs 20.75, find n.

Solution:

As saving is increased by a fixed amount, this will form an AP in which

First term (a) = 5

Common difference (d) = 1.75

an = 20.75

a + (n – 1)d = 20.75

5 + (n – 1) * (1.75) = 20.75

(n – 1) * (1.75) = 20.75 – 5

n – 1 = 15.75/1.75

n = 9 + 1

n = 10

Hence, n = 10

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions: Exercise 5.3

Question 1. Find the sum of the following APs.

(i) 2, 7, 12,â€¦., to 10 terms.

(ii) âˆ’ 37, âˆ’ 33, âˆ’ 29,â€¦, to 12 terms

(iii) 0.6, 1.7, 2.8,â€¦â€¦.., to 100 terms

(iv) 1/15, 1/12, 1/10, â€¦â€¦, to 11 terms

Solution:

(i) Given, 2, 7, 12,â€¦, to 10 terms

For this A.P., we have,

first term, a = 2

common difference, d = a2 âˆ’ a1 = 7âˆ’2 = 5

no. of terms, n = 10

Sum of nth term in AP series is,

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

Substituting the values,

S10 = 10/2 [2(2)+(10 -1)Ã—5]

= 5[4+(9)Ã—(5)]

= 5 Ã— 49 = 245

(ii) Given, âˆ’37, âˆ’33, âˆ’29,â€¦, to 12 terms

For this A.P.,we have,

first term, a = âˆ’37

common difference, d = a2âˆ’ a1

= (âˆ’33)âˆ’(âˆ’37)

= âˆ’ 33 + 37 = 4

no. of terms, n = 12

Sum of nth term in AP series is,

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

Substituting the values,

S12 = 12/2 [2(-37)+(12-1)Ã—4]

= 6[-74+11Ã—4]

= 6[-74+44]

= 6(-30) = -180

(iii) Given, 0.6, 1.7, 2.8,â€¦, to 100 terms

For this A.P.,

first term, a = 0.6

common difference, d = a2 âˆ’ a1 = 1.7 âˆ’ 0.6 = 1.1

no. of terms, n = 100

Sum of nth term in AP series is,

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

S12 = 50/2 [1.2+(99)Ã—1.1]

= 50[1.2+108.9]

= 50[110.1]

= 5505

(iv) Given, 1/15, 1/12, 1/10, â€¦â€¦, to 11 terms

For this A.P.,

first term, a = 1/5

common difference, d = a2 â€“a1 = (1/12)-(1/5) = 1/60

number of terms, n = 11

Sum of nth term in AP series is,

Sn = n/2 [2a + (n â€“ 1) d]

Substituting the values, we have,

= 11/2(2/15 + 10/60)

= 11/2 (9/30)

= 33/20

Question 2. Find the sums given below:

(i) 7+ 10 1/2 + 14 + ……… + 84

(ii) 34 + 32 + 30 + â€¦â€¦â€¦.. + 10

(iii) âˆ’ 5 + (âˆ’ 8) + (âˆ’ 11) + â€¦â€¦â€¦â€¦ + (âˆ’ 230)

Solutions:

(i) Given,

First term, a = 7

nth term, an = 84

Common difference, d = 10 1/2 – 7 = 21/2 – 7 = 7/2

Let 84 be the nth term of this A.P.

Then,

an = a(n-1)d

Substituting these values,

84 = 7+(n â€“ 1)Ã—7/2

77 = (n-1)Ã—7/2

22 = nâˆ’1

n = 23

We know that, sum of n term is;

Sn = n/2 (a + l), l = 84

Sn = 23/2 (7+84)

= (23Ã—91/2) = 2093/2

= 1046 1/2

(ii) Given,

first term, a = 34

common difference, d = a2âˆ’a1 = 32âˆ’34 = âˆ’2

nth term, an= 10

Let 10 be the nth term of this A.P.,

Now,

an = a +(nâˆ’1)d

10 = 34+(nâˆ’1)(âˆ’2)

âˆ’24 = (n âˆ’1)(âˆ’2)

12 = n âˆ’1

n = 13

Sum of n terms is;

Sn = n/2 (a +l), l = 10

= 13/2 (34 + 10)

= (13Ã—44/2) = 13 Ã— 22

= 286

(iii) Given:

First term, a = âˆ’5

nth term, an= âˆ’230

Common difference, d = a2âˆ’a1 = (âˆ’8)âˆ’(âˆ’5)

â‡’d = âˆ’ 8+5 = âˆ’3

Let us assume âˆ’230 be the nth term of this A.P.

Since,

an= a+(nâˆ’1)d

âˆ’230 = âˆ’ 5+(nâˆ’1)(âˆ’3)

âˆ’225 = (nâˆ’1)(âˆ’3)

(nâˆ’1) = 75

n = 76

Sum of n terms, is equivalent to,

Sn = n/2 (a + l)

= 76/2 [(-5) + (-230)]

= 38(-235)

= -8930

Question 3. In an AP

(i) Given a = 5, d = 3, an = 50, find n and Sn.

(ii) Given a = 7, a13 = 35, find d and S13.

(iii) Given a12 = 37, d = 3, find a and S12.

(iv) Given a3 = 15, S10 = 125, find d and a10.

(v) Given d = 5, S9 = 75, find a and a9.

(vi) Given a = 2, d = 8, Sn = 90, find n and an.

(vii) Given a = 8, an = 62, Sn = 210, find n and d.

(viii) Given an = 4, d = 2, Sn = âˆ’ 14, find n and a.

(ix) Given a = 3, n = 8, S = 192, find d.

(x) Given l = 28, S = 144 and there are total 9 terms. Find a.

Solutions:

(i) Given values, we have,

a = 5, d = 3, an = 50

The nth term in an AP,

an = a +(n âˆ’1)d,

Substituting the given values, we have,

â‡’ 50 = 5+(n -1)Ã—3

â‡’ 3(n -1) = 45

â‡’ n -1 = 15

Obtaining the value of n, we get,

â‡’ n = 16

Now, sum of n terms is equivalent to,

Sn = n/2 (a +an)

Sn = 16/2 (5 + 50) = 440

(ii) Given values, we have,

a = 7, a13 = 35

The nth term in an AP,

an = a+(nâˆ’1)d,

Substituting the given values, we have,

â‡’ 35 = 7+(13-1)d

â‡’ 12d = 28

â‡’ d = 28/12 = 2.33

Now, Sn = n/2 (a+an)

Obtaining the final value, we get,

S13 = 13/2 (7+35) = 273

(iii) Given values, we have,

a12 = 37, d = 3

The nth term in an AP,

an = a+(n âˆ’1)d,

Substituting the given values, we have,

â‡’ a12 = a+(12âˆ’1)3

â‡’ 37 = a+33

Obtaining the value of a, we get,

â‡’ a = 4

Now, sum of nth term,

Sn = n/2 (a+an)

= 12/2 (4+37)

Obtaining the final value,

= 246

(iv) Given that,

a3 = 15, S10 = 125

The formula of the nth term in an AP,

an = a +(nâˆ’1)d,

Substituting the given values, we have,

a3 = a+(3âˆ’1)d

15 = a+2d â€¦â€¦â€¦â€¦.. (i)

Also,

Sum of the nth term,

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

S10 = 10/2 [2a+(10-1)d]

125 = 5(2a+9d)

25 = 2a+9d â€¦â€¦â€¦â€¦â€¦.. (ii)

Solving eq (i) by (ii),

30 = 2a+4d â€¦â€¦â€¦. (iii)

And, by subtracting equation (iii) from (ii), we get,

âˆ’5 = 5d

that is,

d = âˆ’1

Substituting in equation (i),

15 = a+2(âˆ’1)

15 = aâˆ’2

a = 17 =

And,

a10 = a+(10âˆ’1)d

a10 = 17+(9)(âˆ’1)

a10 = 17âˆ’9 = 8

(v) Given:

d = 5, S9 = 75

Sum of n terms in AP is,

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

Substituting values, we get,

S9 = 9/2 [2a +(9-1)5]

25 = 3(a+20)

25 = 3a+60

3a = 25âˆ’60

a = -35/3

Also,

an = a+(nâˆ’1)d

Substituting values, we get,

a9 = a+(9âˆ’1)(5)

= -35/3+8(5)

= -35/3+40

= (35+120/3) = 85/3

(vi) Given:

a = 2, d = 8, Sn = 90

Sum of n terms in an AP is,

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

Substituting values, we get,

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

â‡’ 180 = n(4+8n -8) = n(8n-4) = 8n2-4n

Solving the eq, we get,

â‡’ 8n2-4n â€“180 = 0

â‡’ 2n2â€“n-45 = 0

â‡’ 2n2-10n+9n-45 = 0

â‡’ 2n(n -5)+9(n -5) = 0

â‡’ (n-5)(2n+9) = 0

Since, n can only be a positive integer,

Therefore,

n = 5

Now,

âˆ´ a5 = 8+5Ã—4 = 34

(vii) Given:

a = 8, an = 62, Sn = 210

Since, sum of n terms in an AP is equivalent to,

Sn = n/2 (a + an)

210 = n/2 (8 +62)

Solving,

â‡’ 35n = 210

â‡’ n = 210/35 = 6

Now, 62 = 8+5d

â‡’ 5d = 62-8 = 54

â‡’ d = 54/5 = 10.8

(viii) Given :

nth term, an = 4, common difference, d = 2, sum of n terms, Sn = âˆ’14.

Formula of the nth term in an AP,

an = a+(n âˆ’1)d,

Substituting the values, we get,

4 = a+(n âˆ’1)2

4 = a+2nâˆ’2

a+2n = 6

a = 6 âˆ’ 2n â€¦â€¦â€¦â€¦â€¦â€¦â€¦. (i)

Sum of n terms is;

Sn = n/2 (a+an)

-14 = n/2 (a+4)

âˆ’28 = n (a+4)

From equation (i), we get,

âˆ’28 = n (6 âˆ’2n +4)

âˆ’28 = n (âˆ’ 2n +10)

âˆ’28 = âˆ’ 2n2+10n

2n2 âˆ’10n âˆ’ 28 = 0

n2 âˆ’5n âˆ’14 = 0

n2 âˆ’7n+2n âˆ’14 = 0

n (nâˆ’7)+2(n âˆ’7) = 0

Solving for n,

(n âˆ’7)(n +2) = 0

Either n âˆ’ 7 = 0 or n + 2 = 0

n = 7 or n = âˆ’2

Since, we know, n can neither be negative nor fractional.

Therefore, n = 7

From equation (i), we get

a = 6âˆ’2n

a = 6âˆ’2(7)

= 6âˆ’14

= âˆ’8

(ix) Given values are,

first term, a = 3,

number of terms, n = 8

sum of n terms, S = 192

We know,

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

Substituting values,

192 = 8/2 [2Ã—3+(8 -1)d]

192 = 4[6 +7d]

48 = 6+7d

42 = 7d

Solving for d, we get,

d = 6

(x) Given values are,

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

Sum of n terms,

Sn = n/2 (a + l)

Substituting values, we get,

144 = 9/2(a+28)

(16)Ã—(2) = a+28

32 = a+28

Calculating, we get,

a = 4

Question 4. How many terms of the AP. 9, 17, 25 â€¦ must be taken to give a sum of 636?

Solution:

Let us assume that there are n terms of the AP. 9, 17, 25 â€¦

For this A.P.,

We know,

First term, a = 9

Common difference, d = a2âˆ’a1 = 17âˆ’9 = 8

Sum of n terms, is;

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

Substituting the values,

636 = n/2 [2Ã—a+(8-1)Ã—8]

636 = n/2 [18+(n-1)Ã—8]

636 = n [9 +4n âˆ’4]

636 = n (4n +5)

4n2 +5n âˆ’636 = 0

4n2 +53n âˆ’48n âˆ’636 = 0

Solving, we get,

n (4n + 53)âˆ’12 (4n + 53) = 0

(4n +53)(n âˆ’12) = 0

that is,

4n+53 = 0 or nâˆ’12 = 0

On solving,

n = (-53/4) or n = 12

We know,

n cannot be negative or fraction, therefore, n = 12 is the only plausible value.

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.

Solution:

Given:

first term, a = 5

last term, l = 45

Also,

Sum of the AP, Sn = 400

Sum of AP is equivalent to

Sn = n/2 (a+l)

Substituting the values,

400 = n/2(5+45)

400 = n/2(50)

Number of terms, n =16

Since, the last term of AP series is equivalent to

l = a+(n âˆ’1)d

45 = 5 +(16 âˆ’1)d

40 = 15d

Solving for d, we get,

Common difference, d = 40/15 = 8/3

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?

Solution:

Given:

First term, a = 17

Last term, l = 350

Common difference, d = 9

The last term of the AP can be written as;

l = a+(n âˆ’1)d

Substituting the values, we get,

350 = 17+(n âˆ’1)9

333 = (nâˆ’1)9

Solving for n,

(nâˆ’1) = 37

n = 38

Sn = n/2 (a+l)

S38 = 13/2 (17+350)

= 19Ã—367

= 6973

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

Solution:

Given:

Common difference, d = 7

Also,

22nd term, a22 = 149

By the formula of nth term of an AP,

an = a+(nâˆ’1)d

Substituting values, we get,

a22 = a+(22âˆ’1)d

149 = a+21Ã—7

149 = a+147

a = 2 = First term

Sum of n terms,

Sn = n/2(a+an)

S22 = 22/2 (2+149)

= 11Ã—151

= 1661

Question 8. Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18, respectively.

Solution:

Given:

Second term, a2 = 14

Third term, a3 = 18

Also,

Common difference, d = a3âˆ’a2 = 18âˆ’14 = 4

a2 = a+d

14 = a+4

Therefore,

a = 10 = First term

And,

Sum of n terms;

Sn = n/2 [2a + (n â€“ 1)d]

Substituting values,

S51 = 51/2 [2Ã—10 (51-1) 4]

= 51/2 [2+(20)Ã—4]

= 51 Ã— 220/2

= 51 Ã— 110

= 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.

Solution:

Given:

S7 = 49

S17 = 289

Since, we know

Sn = n/2 [2a + (n â€“ 1)d]

Substituting values, we get,

S7= 7/2 [2a +(n -1)d]

S7 = 7/2 [2a + (7 -1)d]

49 = 7/2 [2a + 6d]

7 = (a+3d)

a + 3d = 7 â€¦â€¦â€¦â€¦â€¦â€¦. (i)

Similarly,

S17 = 17/2 [2a+(17-1)d]

Substituting values, we get,

289 = 17/2 (2a +16d)

17 = (a+8d)

a +8d = 17 â€¦â€¦â€¦â€¦â€¦â€¦. (ii)

Solving (i) and (ii),

5d = 10

Solving for d, we get,

d = 2

Now, obtaining value for a, we get,

a+3(2) = 7

a+ 6 = 7

a = 1

Therefore,

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

= n/2[2(1)+(n â€“ 1)Ã—2]

= n/2(2+2n-2)

= n/2(2n)

= n2

(i) an = 3+4n

(ii) an = 9âˆ’5n

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

Solutions:

(i) an = 3+4n

Calculating,

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

Now d =

a2 âˆ’ a1 = 11âˆ’7 = 4

a3 âˆ’ a2 = 15âˆ’11 = 4

a4 âˆ’ a3 = 19âˆ’15 = 4

Hence, ak + 1 âˆ’ ak holds the same value between all pairs of successive terms. Therefore, this is an AP with common difference as 4 and first term as 7.

Sum of nth term is;

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

Substituting the value, we get,

S15 = 15/2[2(7)+(15-1)Ã—4]

= 15/2[(14)+56]

= 15/2(70)

= 15Ã—35

= 525

(ii) an = 9âˆ’5n

Calculating, we get,

a1 = 9âˆ’5Ã—1 = 9âˆ’5 = 4

a2 = 9âˆ’5Ã—2 = 9âˆ’10 = âˆ’1

a3 = 9âˆ’5Ã—3 = 9âˆ’15 = âˆ’6

a4 = 9âˆ’5Ã—4 = 9âˆ’20 = âˆ’11

Common difference, d

a2 âˆ’ a1 = âˆ’1âˆ’4 = âˆ’5

a3 âˆ’ a2 = âˆ’6âˆ’(âˆ’1) = âˆ’5

a4 âˆ’ a3 = âˆ’11âˆ’(âˆ’6) = âˆ’5

Hence, ak + 1 âˆ’ ak holds the same value between all pairs of successive terms. Therefore, this is an AP with common difference as âˆ’5 and first term as 4.

Sum of nth term is;

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

S15 = 15/2[2(4) +(15 -1)(-5)]

Substituting values,

= 15/2[8 +14(-5)]

= 15/2(8-70)

= 15/2(-62)

= 15(-31)

= -465

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions: Exercise 5.3

Question 11. If the sum of the first n terms of an AP is 4n âˆ’ 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, the 10th and the nth terms.

Solution:

Given:

Sn = 4nâˆ’n2

First term can be obtained by putting n=1,

a = S1 = 4(1) âˆ’ (1)2 = 4âˆ’1 = 3

Also,

Sum of first two terms = S2= 4(2)âˆ’(2)2 = 8âˆ’4 = 4

Second term, a2 = S2 âˆ’ S1 = 4âˆ’3 = 1

Common difference, d = a2âˆ’a1 = 1âˆ’3 = âˆ’2

Also,

Nth term, an = a+(nâˆ’1)d

= 3+(n âˆ’1)(âˆ’2)

= 3âˆ’2n +2

= 5âˆ’2n

Therefore, a3 = 5âˆ’2(3) = 5-6 = âˆ’1

a10 = 5âˆ’2(10) = 5âˆ’20 = âˆ’15

Therefore, the sum of first two terms is equivalent to 4. The second term is 1.

And, the 3rd, the 10th, and the nth terms are âˆ’1, âˆ’15, and 5 âˆ’ 2n respectively.

Question 12. Find the sum of first 40 positive integers divisible by 6.

Solution:

The first positive integers that are divisible by 6 are 6, 12, 18, 24 â€¦.

Noticing this series,

First term, a = 6

Common difference, d= 6.

Sum of n terms, we know,

Sn = n/2 [2a +(n â€“ 1)d]

Substituting the values, we get

S40 = 40/2 [2(6)+(40-1)6]

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

= 20(12+234)

= 20Ã—246

= 4920

Question 13. Find the sum of first 15 multiples of 8.

Solution:

The first few multiples of 8 are 8, 16, 24, 32â€¦

Noticing this series,

First term, a = 8

Common difference, d = 8.

Sum of n terms, we know,

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

Substituting the values, we get,

S15 = 15/2 [2(8) + (15-1)8]

= 15/2[6 +(14)(8)]

= 15/2[16 +112]

= 15(128)/2

= 15 Ã— 64

= 960

Question 14. Find the sum of the odd numbers between 0 and 50.

Solution:

The odd numbers between 0 and 50 are 1, 3, 5, 7, 9 â€¦ 49.

Therefore, we can see that these odd numbers are in the form of A.P.

Now,

First term, a = 1

Common difference, d = 2

Last term, l = 49

Last term is equivalent to,

l = a+(nâˆ’1) d

49 = 1+(nâˆ’1)2

48 = 2(n âˆ’ 1)

n âˆ’ 1 = 24

Solving for n, we get,

n = 25

Sum of nth term,

Sn = n/2(a +l)

Substituting these values,

S25 = 25/2 (1+49)

= 25(50)/2

=(25)(25)

= 625

Question 15. A contract on a 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.

Solution:

The given penalties form and A.P. having first term, a = 200 and common difference, d = 50.

By the given constraints,

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

Sum of nth term, we know,

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

Calculating, we get,

S30= 30/2[2(200)+(30 â€“ 1)50]

= 15[400+1450]

= 15(1850)

= 27750

Therefore, the contractor has to pay Rs 27750 as penalty for 30 days delay.

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.

Solution:

Let us assume the cost of 1st prize be Rs. P.

Then, cost of 2nd prize = Rs. P âˆ’ 20

Also, cost of 3rd prize = Rs. P âˆ’ 40

These prizes form an A.P., with common difference, d = âˆ’20 and first term, a = P.

Given that, S7 = 700

Sum of nth term,

Sn = n/2 [2a + (n â€“ 1)d]

Substituting these values, we get,

7/2 [2a + (7 â€“ 1)d] = 700

Solving, we get,

a + 3(âˆ’20) = 100

a âˆ’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?

Solution:

The number of trees planted by the students form an AP, 1, 2, 3, 4, 5â€¦â€¦â€¦â€¦â€¦â€¦..12

Now,

First term, a = 1

Common difference, d = 2âˆ’1 = 1

Sum of nth term,

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

S12 = 12/2 [2(1)+(12-1)(1)]

= 6(2+11)

= 6(13)

= 78

Number of trees planted by 1 section of the classes = 78

Therefore,

Number of trees planted by 3 sections of the classes = 3Ã—78 = 234

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, â€¦â€¦â€¦ as shown in figure. What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take Ï€ = 22/7)

Solution:

We know,

Perimeter of a semi-circle = Ï€r

Calculating ,

P1 = Ï€(0.5) = Ï€/2 cm

P2 = Ï€(1) = Ï€ cm

P3 = Ï€(1.5) = 3Ï€/2 cm

Where, P1, P2, P3 are the lengths of the semi-circles respectively.

Now, this forms a series, such that,

Ï€/2, Ï€, 3Ï€/2, 2Ï€, â€¦.

P1 = Ï€/2 cm

P2 = Ï€ cm

Common difference, d = P2 â€“ P1 = Ï€ â€“ Ï€/2 = Ï€/2

First term = P1= a = Ï€/2 cm

Sum of nth term,

Sn = n/2 [2a + (n â€“ 1)d]

Therefore, Sum of the length of 13 consecutive circles is;

S13 = 13/2 [2(Ï€/2) + (13 â€“ 1)Ï€/2]

Solving, we get,

=  13/2 [Ï€ + 6Ï€]

=13/2 (7Ï€)

= 13/2 Ã— 7 Ã— 22/7

= 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?

[1]

Solution:

The numbers of logs in rows are in the form of an A.P. 20, 19, 18â€¦

Given,

First term, a = 20 and common difference, d = a2âˆ’a1 = 19âˆ’20 = âˆ’1

Let us assume a total of 200 logs to be placed in n rows.

Thus, Sn = 200

Sum of nth term,

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

S12 = 12/2 [2(20)+(n -1)(-1)]

400 = n (40âˆ’n+1)

400 = n (41-n)

400 = 41nâˆ’n2

Solving the eq, we get,

n2âˆ’41n + 400 = 0

n2âˆ’16nâˆ’25n+400 = 0

n(n âˆ’16)âˆ’25(n âˆ’16) = 0

(n âˆ’16)(n âˆ’25) = 0

Now,

Either (n âˆ’16) = 0 or nâˆ’25 = 0

n = 16 or n = 25

By the nth term formula,

an = a+(nâˆ’1)d

a16 = 20+(16âˆ’1)(âˆ’1)

= 20âˆ’15

= 5

And, the 25th term is,

a25 = 20+(25âˆ’1)(âˆ’1)

= 20âˆ’24

= âˆ’4

Therefore, 200 logs can be placed in 16 rows and the number of logs in the 16th row is 5, since the number of logs can’t be negative as in case of 25th term.

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 meters) run by a competitor is 2Ã—5+2Ã—(5+3)]

Solution:

The distances of potatoes from the bucket are 5, 8, 11, 14â€¦, which form an AP.

Now, we know that  the 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 w.r.t distances of potatoes is equivalent to,

10, 16, 22, 28, 34,â€¦â€¦â€¦.

We get, a = 10 and d = 16âˆ’10 = 6

Sum of nth term, we get,

S10 = 12/2 [2(20)+(n -1)(-1)]

= 5[20+54]

= 5(74)

Solving we get,

= 370

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

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions: Exercise 5.4

[Hint: Find n for an < 0]

Solution:

Given the AP series is 121, 117, 113, . . .,

Here, first term, a = 121 and Common difference, d = 117-121= -4

nth term formula, an = a+(n âˆ’1)d

Hence,

an = 121+(nâˆ’1)(-4)

= 121-4n+4

=125-4n

To find the first negative term of the series, an < 0 Therefore,

125-4n < 0

125 < 4n

n>125/4

n>31.25

Therefore, the first negative term of the series is 32nd term (n=32).

which will be, a32 =125-4(32) = -3

Question 2. The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP.

Solution:

nth term formula, an = a+(n âˆ’1)d

Third term, a3 = a+(3 -1)d

a3 = a + 2d

And Seventh term, a7= a+(7-1)d

a7 = a + 6d

According to given conditions,

a3 + a7 = 6 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.(i)

a3 Ã— a7 = 8 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..(ii)

Substituting the values in eqn. (i), we get,

a+2d +a+6d = 6

2a+8d = 6

a+4d=3

a = 3â€“4d â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦(iii)

Now substituting the values in eqn. (ii), we get,

(a+2d)Ã—(a+6d) = 8…………………..(iv)

Putting the value of a from eqn. (iii) in eqn. (iv), we get,

(3â€“4d +2d)Ã—(3â€“4d+6d) = 8

(3 â€“2d)Ã—(3+2d) = 8

32 â€“ (2d)2 = 8         (using the identity, (a+b)(a-b) = a2-b2)

9 â€“ 4d2 = 8

4d2 = 9-8 = 1

d = âˆš(1/4)

d = Â±1/2

d = 1/2 or -1/2

So now, if

d = 1/2,

a = 3 â€“ 4d = 3 â€“ 4(1/2) = 3 â€“ 2 = 1

and if d = -1/2,

a = 3 â€“ 4d = 3 â€“ 4(-1/2) = 3+2 = 5

Sum of nth term of AP is: Sn = n/2 [2a +(n â€“ 1)d]

So, when a = 1 and d=1/2

Then, the sum of first 16 terms are;

S16 = 16/2 [2 +(16-1)1/2] = 8(2+15/2)

S16 = 76

And when a = 5 and d= -1/2

Then, the sum of first 16 terms are;

S16 = 16/2 [2.5+(16-1)(-1/2)] = 8(5/2)

S16 = 20

[Hint : Number of rungs = 250/25 + 1]

Solution:

As it is given that,

Distance between the rungs of the ladder is 25 cm.

Distance between the top rung and bottom rung of the ladder will be (in cm) = 2 Â½ Ã— 100 = 250 cm

Hence, the total number of rungs = 250/25 + 1 = 11

As we can observe here, that, the ladder has rungs in decreasing order from top to bottom. Thus, the rungs are decreasing in an order of AP.

So, According to given condition

First term, a = 45

Last term, l = 25

Number of terms, n = 11

And the length of the wood required for the rungs will be equal to the sum of the terms of AP series formed.

Now, as we know, sum of nth terms is equal to, Sn= (n/2)(a+ l)

Sn= 11/2(45+25) = 11/2(70) = 385 cm

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

[Hint : Sx â€“ 1 = S49 â€“ Sx]

Solution:

According to the given statement,

Row houses are numbered in a row that are in the form of AP, which is as follows:

 1 2 3 ….. ….. …. 49

Here, according to the given condition,

First term, a = 1

Common difference, d=1

and last term = l

Let us say the number of xth houses can be represented as: Sx â€“ 1 = S49 â€“ Sx

Where Sx represents to sum of xth houses.

(Value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it)

As, Sum of nth term of AP, Sn= n/2[a+l]

Sum of number of houses beyond x house = Sx-1

Sx-1 = (x-1)/2[1+(x-1)]

Sx-1 = x(x-1)/2 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦(i)

By the given condition, we can write,

S49 â€“ Sx = {49/2[1+(49)}â€“{x/2(1+x)}

= 25(49) â€“ x(x + 1)/2 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.(ii)

As eqn. (i) and (ii) are equal, So

x(x-1)/2 = 25(49) â€“ x(x-1)/2

x(x-1) = 25(49)

x = Â±35

As we know, the number of houses cannot be a negative number.

Hence, the value of x is 35.

[Hint : Volume of concrete required to build the first step = 1/4 Ã— 1/2 Ã— 50 m3]

Solution:

As we can see from the given figure,

1st step is Â¼ m high, Â½ m wide and 50 m long

2nd step is (Â¼+Â¼ = Â½ m) high, Â½ m wide and 50 m long and,

3rd step is (3Ã—Â¼ = 3/4 m) high, Â½ m wide and 50 m long.

and so on

Hence, we can conclude that the height of step increases by Â¼ m each time when width and length is Â½ m and 50 m respectively.

So, the height of steps forms a series AP in such a way that;

Â¼, Â½ , Â¾, 1, 5/4, â€¦â€¦..

Volume of steps = Volume of Cuboids

= Length Ã— Breadth Ã— Height

Now,

Volume of concrete required to build the first step = Â¼ Ã—Â½ Ã—50 = 25/4

Volume of concrete required to build the second step =Â½ Ã—Â½Ã—50 = 25/2

Volume of concrete required to build the second step = Â¾ Ã— Â½ Ã—50 = 75/2

Volume of steps = Â½ Ã— 50 Ã— (Â¼ + 2/4 + 3/4 + 4/4 + 5/4 + …….) …………………….(1)

Now, we can see the height of concrete required to build the steps, are in AP series;

Thus, applying the AP series concept to the height,

First term, a = 1/4

Common difference, d = 2/4-1/4 = 1/4

n = 15

As, the sum of n terms is : Sn = n/2[2a+(n-1)d]

Sn = 15/2(2Ã—(1/4 )+(15 -1)1/4)

Sn = 15/2 (4)

Sn = 30

Hence, up solving eqn. (1), we get

Volume of steps = Â½ Ã— 50 Ã— Sn

= Â½ Ã— 50 Ã— 30

= 750 m3

Hence, the total volume of concrete required to build the terrace is 750 m3.

Key Features of NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions:

• NCERT Solutions are created for each chapter including NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions.
• These solutions provide in-depth solutions to the problems encountered by students in their NCERT Class 10 Maths textbook.
• These solutions are very accurate and comprehensive, which can help students of Class 10 prepare for any academic as well as competitive exams.

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NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progression- FAQs

Q1: Why is it important to learn arithmetic progressions?

Learning arithmetic progressions is important for Class 10 students for several reasons. Arithmetic progressions serve as a foundation for more advanced mathematical concepts that students will encounter in higher grades. By understanding arithmetic progressions, students develop skills in recognizing patterns, understanding the relationship between terms, and finding the common difference.

Q2: What topics are covered in NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions?

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions covers topics such sum of series till n terms, calculating first, last and nth term of series etc.

Q3: How can NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions help me?

NCERT Solutions for Class 10 Maths Chapter 5 â€“ Arithmetic Progressions can help you solve the NCERT exercise without any limitations. If you are stuck on a problem you can find its solution in these solutions and free yourself from the frustration of being stuck on some question.

Q4: How many exercises are there in NCERT Class 10 Maths Chapter 5 Arithmetic Progressions?

There are 4 exercises in the Class 10 Maths Chapter 5 â€“ Arithmetic Progressions which covers all the important topics and sub-topics.