NCERT Solutions for Class 10 Maths Chapter 5 – Arithmetic Progressions

NCERT Solutions for Class 10 Maths Chapter 5 – Arithmetic Progressions. Furthermore, here we’ve provided you with the latest solution for Class 10 CBSE NCERT Maths Chapter 5 Arithmetic Progressions. As a result here you’ll find solutions to all the exercises. This NCERT Class 10 solution will help you to score good marks in your exam.

What is Arithmetic Progression?

An arithmetic progression is a sequence of numbers such that the difference between two consecutive terms is constant. In this chapter of ncert class 10 there are 4 exercises. You’ll find a detailed solution of all the questions here.

Arithmetic Progression Formulae

Here you’ll find the list of all formulae which we are going to use to solve the entire chapter.
1) t_n = a + (n-1) d
t_n is the nth term which we have to find
where ‘a’ is the constant term,
‘n’ is the number of terms and
‘d’ is the common difference of the AP.

2) S = n/2 [2a + (n – 1)d]
This formula is used to find the sum of the first ‘n’ terms of an arithmetic sequence

NCERT Solution for Class 10 Maths Chapter 5 – Arithmetic Progression Pdf Download

Students can refer to our solution for NCERT Class 10 Maths Chapter 5 – Arithmetic Progression. The Chapter 5 Solution of NCERT will help students prepare for the exams and easily crack the exam. Below we’ve provided you with the exercise wise latest solution.

NCERT Class 10 Chapter 5 Arithmetic Progression Exercise wise solution

Exercise 5.1 – Page 99 of NCERT
Exercise 5.2 – Page 105 of NCERT
Exercise 5.3 – Page 112 of NCERT
Exercise 5.4 – Page 115 of NCERT

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

Here you’ll find NCERT Chapter 5 Arithmetic Progressions Exercise 5.1 Solution.
Exercise 5.1: Solutions of Questions on Page Number: 99.

Q1. In which of the following situations, does the list of numbers involved make an arithmetic progression and why?
(i) The taxi fare after each km when the fare is Rs 15 for the first km and Rs 8 for each additional km.
(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.
(iii) 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.
(iv) The amount of money in the account every year, when Rs 10000 is deposited at compound interest at 8% per annum.
Answer:
(i) It can be observed that Taxi fare for 1^stkm = 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.
Let the initial volume of air in a cylinder be V lit. In each stroke, the vacuum pump removes 1/4 of the air remaining in the cylinder at a time. In other words, after every stroke, only 1 – 1/4 = 3/4 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.

(ii) 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.

(iv) 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.

Q2.

Write first four terms of the A.P. when the first term a and the common differenced are given as follows
(i) a = 10, d = 10
(ii) a = – 2, d = 0
(iii) a = 4, d = – 3
(iv) a = – 1, d = 1/2
(v) a = – 1.25, d = – 0.25

Answer:
(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.

(ii) 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.

(iii) a = 4, d = – 3
Let the series be a₁, a₂, a₃, a₄, …
a₁ = a = -2
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.

(iv) a = -1 d = 1/2
Let the series be a₁, a₂, a₃, a₄, …

Clearly, the series will be

First four terms of this A.P. will be

(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.

Q3: 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) 1/3, 5/3, 9/3, 13/3 …
(iv) 0.6, 1.7, 2.8, 3.9 …

Answer:
(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) 1/3, 5/3, 9/3, 13/3 …
Here first term a = 1/3
Common difference, d = Second term – First term
= 5/3 – 1/3 = 4/3

(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

Q4. 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) 2, 5/2, 3, 7/2 …
(iii) -1.2, -3.2, -5.2, -7.2 …
(iv) -10, -6, -2, 2 …
(v)
(vi) ) 0.2, 0.22, 0.222, 0.2222 …
(vii) 0, – 4, – 8, – 12 …
(viii) -1/2, -1/2, -1/2, -1/2 …
(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 …

Answer:
(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) 2, 5/2, 3, 7/2 …
It can be observed that

i.e., a_k+1 – a_k is same every time.
Therefore, d = 1/2 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., a_k+1 – a_k is 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., a_k+1 – a_k is 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., a_k+1 – a_k 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., a_k+1 – a_k 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., a_k+1 – a_k is same every time.
Therefore, d = – 4
The given numbers are in A.P.
a5 = 2 + (-4) = -2
a6 = -2 + (-4) = -6
a7 = -6 + (-4) = 10

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Exercise 5.2

Here you’ll find NCERT Chapter 5 Arithmetic Progressions Exercise 5.2 Solution.
Exercise 5.2: Solutions of Questions on Page Number: 105.

Q1. Fill in the blanks in the following table, given that a is the first term, d the common difference and an the n^thterm 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	…..

Answer:

(i) a = 7, d = 3, n = 8, a_n = ?
We know that,
For an A.P. a_n = a + (n – 1) d
= 7 + (8 – 1) 3
= 7 + (7) 3
= 7 + 21 = 28
Hence,a_n = 28

(ii) Given that
a = – 18, n = 10, a_n = 0, d = ?
We know that,
a_n = a + (n – 1) d
0 = – 18 + (10 – 1) d
18 = 9d
d = 2
Hence, common difference, d = 2

(iii) Given that
d = – 3, n = 18, a_n = – 5
We know that,
a_n = a + (n – 1) d
– 5 = a + (18 – 1) ( -3)
– 5 = a + (17) ( -3)
– 5 = a – 51
a = 51 – 5 = 46
Hence, a = 46

(iv) a = – 18.9, d = 2.5, a_n = 3.6, n = ?
We know that,
a_n = a + (n – 1) d
3.6 = – 18.9 + (n – 1) 2.5
3.6 + 18.9 = (n – 1) 2.5
22.5 = (n – 1) 2.5
(n-1) = 22.5/2.5
n – 1 = 9
n = 10
Hence n = 10.

(v) a = 3.5, d = 0, n = 105, a_n = ?
We know that,
a_n = a + (n – 1) d
a_n = 3.5 + (105 – 1) 0
a_n = 3.5 + 104 × 0
a_n = 3.5
Hence, an = 3.5.

Q2: Choose the correct choice in the following and justify
(i) 30^thterm of the A.P: 10, 7, 4, …, is
A) 97 B) 77 C) -77, D) -87

(ii) 11th term of the A.P. -3, -1/2, 2 … is
A) 28 B) 22 C) -38 D) -48 x 1/2

Answer
(i) Given that
A.P. 10, 7, 4, …
First term, a = 10
Common difference, d = a₂ – a₁ = 7 – 10
= – 3
We know that,
a_n = a + (n – 1) d
a₃₀ = 10 + (30 – 1) ( – 3)
a₃₀ = 10 + (29) ( – 3)
a₃₀ = 10 – 87 = – 77
Hence, the correct answer is C) -77

(ii) Given that, A.P. First term a = – 3
Common difference, d = a₂ – a₁

We know that,

Hence, the answer is B) 22

Q3: In the following APs find the missing term in the boxes.
(i)
(ii)
(iii)
(iv)
(v)

Answers
(i) For this A.P.,
a = 2
a₃ = 26
We know that, an = a + (n – 1) d a₃ = 2 + (3 – 1) d
26 = 2 + 2d
24 = 2d
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)

We know that

Therefore, the missing terms are 13/2 and 8 respectively.

(iv) For this A.P.,
a = – 4 and a₆ = 6
We know that, an = a + (n – 1) d
a₆ = a + (6 – 1) d
6 = – 4 + 5d
10 = 5d
d = 2
a₂ = a + d = – 4 + 2 = – 2
a₃ = a + 2d = – 4 + 2 (2) = 0
a₄ = a + 3d = – 4 + 3 (2) = 2
a₅ = a + 4d = – 4 + 4 (2) = 4
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
a₂ = a + (2 – 1) d 38 = a + d … (1)
a₆ = a + (6 – 1) d
-22 = a + 5d … (2)
On subtracting equation (1) from (2), we obtain
-22 – 38 = 4d
-60 = 4d d = – 15
a = a₂ – d = 38 – ( – 15) = 53
a₃ = a + 2d = 53 + 2 ( – 15) = 23
a₄ = a + 3d = 53 + 3 ( – 15) = 8
a₅ = a + 4d = 53 + 4 ( – 15) = – 7
Therefore, the missing terms are 53, 23, 8, and – 7 respectively.

Q4: Which term of the A.P. 3, 8, 13, 18, … is 78?
Answer:
3, 8, 13, 18, …
For this A.P.,
a = 3
d = a₂ – a₁ = 8 – 3 = 5
Let n^thterm of this A.P. be 78.
a_n = a + (n – 1) d
78 = 3 + (n – 1) 5
75 = (n – 1) 5
(n – 1) = 15
n = 16
Hence, 16^thterm of this A.P. is 78.

Q5: Find the number of terms in each of the following A.P.
(i) 7, 13, 19, …, 205
(ii)

Answer:
(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
a_n = a + (n – 1) d
Therefore, 205 = 7 + (n – 1) 6
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, a_n = – 47 and we know that,

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

Q6: Check whether – 150 is a term of the A.P. 11, 8, 5, 2, …
Answer:
For this A.P.,
a = 11
d = a₂ – a₁ = 8 – 11 = – 3
Let – 150 be the n^thterm of this A.P
We know that,
a₁ = a+ (n-1)d
-150-11+(n-1) (-3)
-150=11-3n+3
-164=-3n
n =164/3
Clearly, n is not an integer.
Therefore, -150 is not a term of this A.P.

Q7: Find the 31^stterm of an A.P. whose 11^thterm is 38 and the 16^thterm is 73.
Answer:
Given that,
a₁₁ = 38
a₁₆ = 73
We know that,
an = a + (n – 1) d
a₁₁ = a + (11 – 1) d
38 = a + 10d (1)
Similarly,
a₁₆ = a + (16 – 1) d
73 = a + 15d (2)
On subtracting (1) from (2),
we obtain 35 = 5d
d = 7
From equation (1),
38 = a + 10 x (7)
38 – 70 = a
a = -32
a₃₁= a + (31 – 1) d
= -32 + 30 (7)
= -32 + 210
= 178
Hence, 31^stterm is 178.

Q8: An A.P. consists of 50 terms of which 3^rdterm is 12 and the last term is 106. Find the 29^thterm.
Answer:
Given that,
a₃ = 12
a₅₀ = 106
We know that,
an = a + (n – 1) d
a₃ = a + (3 – 1) d
12 = a + 2d … (i)
Similarly, a₅₀ = a + (50 – 1) d
106 = a + 49d …. (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, 29^thterm is 64.

Q9: If the 3^rdand the 9^thterms of an A.P. are 4 and – 8 respectively. Which term of this A.P. is zero.
Answer:
Given that, a₃ = 4
a₉ = -8
We know that
a_n = a + (n – 1) d
a₃ = a + (3 – 1) d
4 = a + 2d … (i)
a₉ = a + (9 – 1) d
-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 n^thterm of this A.P. be zero.
an = a + (n – 1) d
0 = 8 + (n – 1) (-2)
0 = 8 – 2n + 2
2n = 10
n = 5
Hence, 5^thterm of this A.P. is 5.

Q10: If 17^thterm of an A.P. exceeds its 10^thterm by 7. Find the common difference.
Answer:
We know that,

For an A.P., an = a + (n – 1) d
a17 = a + (17 – 1) d
a17 = a + 16d
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.

Q11. Which term of the A.P. 3, 15, 27, 39, … will be 132 more than its 54^thterm?
Answer:
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 n^thterm be 771.
a_n = a + (n – 1) d
771 = 3 + (n – 1) 12
768 = (n – 1) 12
(n – 1) = 64
n = 65
Therefore, 65^thterm was 132 more than 54^thterm.

Alternatively,
Let n^thterm be 132 more than 54^thterm.

Q12: Two APs have the same common difference. The difference between their 100^thterm is 100, what is the difference between their 1000^thterms?

Answer: 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.,
a₁₀₀ = a₁ + (100 – 1) d
=a₁ + 99d
a₁₀₀₀ = a1 + (1000 – 1) d a1000
= a₁ + 999d
For second A.P.,
a100 = a₂ + (100 – 1) d
=a₂ + 99d
a1000 = a₂ + (1000 – 1) d
= a₂ + 999d
Given that, difference between 100^thterm of these A.P.s = 100
Therefore, (a₁ + 99d) – (a2 + 99d) = 100
a₁ – a₂ = 100 …(1)
Difference between 1000^thterms of these A.P.s (a1 + 999d) – (a2 + 999d) = a1 – a2
From equation (1),
This difference, a₁ – a₂ = 100
Hence, the difference between 1000^thterms of these A.P. will be 100.

Q13: How many three digit numbers are divisible by 7
Answer:
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
a_n = 994
n = ?
a_n = a + (n – 1) d
994 = 105 + (n – 1) 7
889 = (n – 1) 7
(n – 1) = 127
n = 128
Therefore, 128 three-digit numbers are divisible by 7.

Q14: How many multiples of 4 lie between 10 and 250?
Answer:
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 n^thterm of this A.P.

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

Q15: For what value of n, are the n^th terms of two APs 63, 65, 67, and 3, 10, 17, … equal.
Answer:
63, 65, 67, …
a = 63
d = a2 – a1 = 65 – 63 = 2
n^th term of this A.P. = an = a + (n – 1) d
a_n= 63 + (n – 1) 2 = 63 + 2n – 2
a_n = 61 + 2n (1)
3, 10, 17, …
a = 3
d = a₂ – a₁ = 10 – 3 = 7
n^th term of this A.P. = 3 + (n – 1) 7
a_n= 3 + 7n – 7
a_n= 7n – 4 (2)
It is given that, n^th 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, 13^th terms of both these A.P.s are equal to each other.

Q16: Determine the A.P. whose third term is 16 and the 7^th term exceeds the 5^th term by 12.
Answer:
=a₃ = 16
a + (3 – 1) d = 16
a + 2d = 16 … (1)
a7 – a5 = 12
[a+ (7 – 1) d] – [a + (5 – 1) d]= 12 (a + 6d) – (a + 4d) = 12
2d = 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, …

Q17: Find the 20^th term from the last term of the A.P. 3, 8, 13, …, 253.
Answer:
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, 5
For this A.P.,
a = 253
d = 248 – 253 = -5
n = 20
a₂₀ = a + (20 – 1) d a20 = 253 + (19) (-5)
a₂₀ = 253 – 95
a = 158.
Therefore, 20^th term from the last term is 158.

Q18: The sum of 4^th and 8^th terms of an A.P. is 24 and the sum of the 6^th and 10^th terms is 44. Find the first three terms of the A.P.
Answer:
We know that, an = a + (n – 1) d
a4=a + (4 – 1) d
a4 = a + 3d
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 … (i)
a6 + a10 = 44
a + 5d + a + 9d = 44
2a + 14d = 44
a + 7d = 22 … (ii)
On subtracting equation (i) from (ii),
we obtain 2d = 22 – 12
2d = 10
d = 5
From equation (i), 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.

Q19: Ramkali saved Rs 5 in the first week of a year and then increased her weekly saving by Rs 1.75. If in the n^thweek, her week, her weekly savings become Rs 20.75, find n.
Answer:
Given that, a = 5
d = 1.75
a_n = 20.75
n = ?
a_n = a + (n – 1) d

n – 1 = 9
n = 10
Hence, n is 10.

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

Here you’ll find NCERT Chapter 5 Arithmetic Progressions Exercise 5.3 Solution.
Exercise 5.3: Solutions of Questions on Page Number: 112.

Q1: 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

Answer:
(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.
a = 1/15
n = 11
d = a₂– a₁ = 1/12 – 1/15
= (5 – 4)/60 = 1/60.
We know that

Q2:

Find the sums given below

(i) 7 + 10/2 + 14 +     + 84
(ii) 34 + 32 + 30 +        + 10
(iii) – 5 + ( – 8) + ( – 11) +   + ( – 230)

Answer:
(i) 7 + 10/2 + 14 + + 84
For this A.P.,
a = 7
l = 84

Let 84 be the n^th term of this A.P.

l = a + (n – 1)d

22 = n – 1
n = 23

(ii) 34 + 32 + 30 + + 10
For this A.P.,
a = 34
d = a₂ – a₁ = 32 – 34 = – 2
l = 10
Let 10 be the n^th term of this A.P.
l = a + (n – 1) d
10 = 34 + (n – 1) ( – 2)
-24 = (n – 1) ( – 2)
12 = n – 1
n = 13

(iii)( – 5) + ( – 8) + ( – 11) + + ( – 230)

For this A.P.,
a = – 5
l = – 230
d = a₂ – a₁ = ( – 8) – ( – 5)
= – 8 + 5 = – 3
Let – 230 be the n^th term of this A.P.
l = a + (n – 1)d
-230 = – 5 + (n – 1) ( – 3)
-225 = (n – 1) ( – 3)
(n – 1) = 75
n = 76

Q3:

In an AP

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

1) Given that, a = 5, d = 3, a_n = 50 As a_n = a + (n – 1)d,
∴ 50 = 5 + (n – 1)3
45 = (n – 1)3
15 = n – 1
n = 16

(ii) 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
d = 7/3

(iii) Given that, a12 = 37, d = 3
As a_n = a + (n – 1)d,
a₁₂ = a + (12 – 1)3
37 = a + 33
a = 4

(iv) Given that, a₃ = 15, S₁₀ = 125
As a_n = 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.

(v) Given that, d = 5, S₉ = 75

25 = 3(a + 20)
25 = 3a + 60
3a = 25 – 60
a = -35/3
a_n = a + (n – 1)d
a₉ = a + (9 – 1) (5)

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

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 n = -18/4 = -9/2
However, n can neither be negative nor fractional.
Therefore, n = 5.

Q4: 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 = 9d = 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
n = -53/4 or n = 12
n cannot be negative nor fractional, therefore, n = 12 only.

Q5: 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.
Answer:
Given that, a = 5
l = 45
Sn = 400

n = 16
l = a + (n – 1) d
45 = 5 + (16 – 1) d
40 = 15d
d = 40/15 = 8/3

Q6: 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?
Answer:
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.

Q7: Find the sum of first 22 terms of an AP in which d = 7 and 22^nd term is 149.
Answer:
d = 7
a₂₂ = 149
S₂₂ = ?
an = a + (n – 1)d
a₂₂ = a + (22 – 1)d
149 = a + 21 × 7
149 = a + 147
a = 2

Q8: Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
Answer:
Given that,
a₂ = 14
a₃ = 18
d = a₃ – a₂ = 18 – 14 = 4
a₂ = a + d
14 = a + 4
a = 10

=5610

Q9: 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.
Answer:
Given that,
S₇ = 49
S₁₇ = 289

7 = (a + 3d)
a + 3d = 7 … (i)

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

Q10: Show that a₁, a₂ … , a_n , … form an AP where an is defined as below
(i) a_n = 3 + 4n
(ii) a_n = 9 – 5n
Also find the sum of the first 15 terms in each case.

Answer:
a_n = 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., a_{k + 1} – a_k is same every time.
Therefore, this is an AP with common difference as 4 and first term as 7.

15 × 35
=525

(ii) a_n = 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., a_{k + 1} – a_k is same every time.
Therefore, this is an A.P. with common difference as – 5 and first term as 4.

= -465

Q11: 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 3^rd, the 10^th and the n^th terms.
Answer:
Given that,
S_n = 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
a_n = 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. 3^rd, 10^th, and n^th terms are -1, -15, and 5 – 2n respectively.

Q12: Find the sum of first 40 positive integers divisible by 6
Answer:
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

Q13: Find the sum of first 15 multiples of 8
Answer:
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

Q14: Find the sum of the odd numbers between 0 and 50.
Answer:
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

Q15: 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.
Answer:
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.

Q16: 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.
Answer:
Let the cost of 1^st prize be P.
Cost of 2^nd prize = P – 20
And cost of 3^rd 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.

Q17: 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?

Answer:
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.

Q18: 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. (Take π = 22/7)

Answer: Semi-perimeter of circle = πr
I₁ = π(0.5) = π/2
I₂ = π(1) = π cm
I₃ = π(1.5) = 3π/2
Therefore, I₁, I₂, I₃ ,i.e. the lengths of the semi-circles are in an A.P.,
π/2, π, 3π/2, 2π,………..

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.

Q19: 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?

Answer:
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 – 41n + 400 = 0
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
a_n = 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 16^th row is 5. However, the number of logs in 25^th row is negative, which is not possible. Therefore, 200 logs can be placed in 16 rows and the number of logs in the 16^th row is 5.

Q20: 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 x 5 + 2 x (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.

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

Here you’ll find NCERT Chapter 5 Arithmetic Progressions Exercise 5.4 Solution.
Exercise 5.4: Solutions of Questions on Page Number: 115.

Q1: Which term of the A.P. 121, 117, 113 … is its first negative term? [Hint: Find n for an < 0]
Answer:
Given A.P. is 121, 117, 113 …
a = 121
d = 117 – 121 = – 4
a_n = 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, 32^nd term will be the first negative term of this A.P.

Q2: 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.
Answer:
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)

Q3: 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 are 2×1/2 m apart, what is the length of the wood required for the rungs? [Hint: number of rungs = 250/25]

Answer: It is given that the rungs are 25 cm apart and the top and bottom rungs are 2×1/2 m 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.

Q4: 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].
Answer:
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 x^th house was like this. We know that,
Sum of n terms in an A.P

Sum of number of houses preceding x^th house = S_{x – 1}

Sum of number of houses following x^th house = S₄₉ – S_x

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.

Q5: 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 1/4 m and a tread of 1/4m (See figure) calculate the total volume of concrete required to build the terrace.

Answer:

From the figure, it can be observed that
1st step is 1/2m wide
2nd step is 1m wide
3rd step is 3/2 m wide.
Therefore, the width of each step is increasing by 1/2 m each time whereas their height 1/4 m and length 50 m remains the same.
Therefore, the widths of these steps are

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.

We hope the NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions helps you. If you have any query regarding chapter 5. Drop a comment below and we will get back to you at the earliest.