Chapter 10 : Arithmetic Progression  Selina Solutions for Class 10 Maths ICSE
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Chapter 10  Arithmetic Progression Excercise Ex. 10(A)
Which of the following sequences are in arithmetic progression?
(i) 2, 6, 10, 14,
(ii) 15, 12, 9, 6,
(iii) 5, 9, 12, 18,
(iv)
The n^{th} term of sequence is (2n  3), find its fifteenth term.
If the p^{th} term of an A.P. is (2p + 3), find the A.P.
Find the 24^{th} term of the sequence:
12, 10, 8, 6,……
Find the 30^{th} term of the sequence:
Is 402 a term of the sequence :
8, 13, 18, 23,………….?
How many terms are there in the series :
(i) 4, 7, 10, 13, …………, 148?
(ii) 0.5, 0.53, 0.56, ……………, 1.1?
Which term of the A.P. 1 + 4 + 7 + 10 + ………. is 52?
If 5^{th} and 6^{th} terms of an A.P are respectively 6 and 5. Find the 11^{th} term of the A.P
If t_{n} represents n^{th} term of an A.P, t_{2} + t_{5}  t_{3} = 10 and t_{2} + t_{9} = 17, find its first term and its common difference.
Find the 10^{th} term from the end of the A.P. 4, 9, 14,…….., 254
Determine the arithmetic progression whose 3^{rd} term is 5 and 7^{th} term is 9.
Find the 31^{st} term of an A.P whose 10^{th} term is 38 and 16^{th} term is 74.
Which term of the services :
21, 18, 15, …………. is  81?
Can any term of this series be zero? If yes find the number of term.
An A.P. consists of 60 terms, If the first and the last terms be 7 and 125 respectively, find the 31^{st} term.
For a given A.P.,
Number of terms, n = 60
First term, a = 7
Last term, l = 125
⇒ t_{60} = 125
⇒ a + 59d = 125
⇒ 7 + 59d = 125
⇒ 59d = 118
⇒ d = 2
Hence, t_{31} = a + 30d = 7 + 30(2) = 7 + 60 = 67
The sum of the 4^{th} and the 8^{th} terms of an A.P. is 24 and the sum of the 6^{th} and the 10^{th} terms of the same A.P. is 34. Find the first three terms of the A.P.
Let 'a' be the first term and 'd' be the common difference of the given A.P.
t_{4} + t_{8} = 24 (given)
⇒ (a + 3d) + (a + 7d) = 24
⇒ 2a + 10d = 24
⇒ a + 5d = 12 ….(i)
And,
t_{6} + t_{10} = 34 (given)
⇒ (a + 5d) + (a + 9d) = 34
⇒ 2a + 14d = 34
⇒ a + 7d = 17 ….(ii)
Subtracting (i) from (ii), we get
2d = 5
If the third term of an A.P. is 5 and the seventh term is 9, find the 17^{th} term.
Let 'a' be the first term and 'd' be the common difference of the given A.P.
Now, t_{3} = 5 (given)
⇒ a + 2d = 5 ….(i)
And,
t_{7} = 9 (given)
⇒ a + 6d = 9 ….(ii)
Subtracting (i) from (ii), we get
4d = 4
⇒ d = 1
⇒ a + 2(1) = 5
⇒ a = 3
Hence, 17^{th} term = t_{17} = a + 16d = 3 + 16(1) = 19
Chapter 10  Arithmetic Progression Excercise Ex. 10(B)
In an A.P., ten times of its tenth term is equal to thirty times of its 30^{th} term. Find its 40^{th} term.
How many twodigit numbers are divisible by 3?
Which term of A.P. 5, 15, 25 ………… will be 130 more than its 31^{st} term?
Find the value of p, if x, 2x + p and 3x + 6 are in A.P
If the 3^{rd} and the 9^{th} terms of an arithmetic progression are 4 and 8 respectively, Which term of it is zero?
How many threedigit numbers are divisible by 87?
For what value of n, the n^{th} term of A.P 63, 65, 67, …….. and n^{th} term of A.P. 3, 10, 17,…….. are equal to each other?
Determine the A.P. Whose 3^{rd} term is 16 and the 7^{th} term exceeds the 5^{th} term by 12.
If numbers n  2, 4n  1 and 5n + 2 are in A.P. find the value of n and its next two terms.
Determine the value of k for which k^{2 }+ 4k + 8, 2k^{2} + 3k + 6 and 3k^{2} + 4k + 4 are in A.P
If a, b and c are in A.P show that:
(i) 4a, 4b and 4c are in A.P
(ii) a + 4, b + 4 and c + 4 are in A.P.
An A.P consists of 57 terms of which 7^{th} term is 13 and the last term is 108. Find the 45^{th} term of this A.P.
4^{th} term of an A.P is equal to 3 times its first term and 7^{th} term exceeds twice the 3^{rd} time by I. Find the first term and the common difference.
The sum of the 2^{nd} term and the 7^{th} term of an A.P is 30. If its 15^{th} term is 1 less than twice of its 8^{th} term, find the A.P
In an A.P, if m^{th} term is n and n^{th} term is m, show that its r^{th} term is (m + n  r)
Which term of the A.P 3, 10, 17, ………. Will be 84 more than its 13^{th} term?
Chapter 10  Arithmetic Progression Excercise Ex. 10(C)
Find the sum of the first 22 terms of the A.P.: 8, 3, 2, ………..
How many terms of the A.P. :
24, 21, 18, ……… must be taken so that their sum is 78?
Find the sum of 28 terms of an A.P. whose n^{th} term is 8n  5.
Find the sum of all odd natural numbers less than 50
Find the sum of first 12 natural numbers each of which is a multiple of 7.
Find the sum of first 51 terms of an A.P. whose 2^{nd} and 3^{rd} terms are 14 and 18 respectively.
The sum of first 7 terms of an A.P is 49 and that of first 17 terms of it is 289. Find the sum of first n terms
The first term of an A.P is 5, the last term is 45 and the sum of its terms is 1000. Find the number of terms and the common difference of the A.P.
Find the sum of all natural numbers between 250 and 1000 which are divisible by 9.
The first and the last terms of an A.P. are 34 and 700 respectively. If the common difference is 18, how many terms are there and what is their sum?
In an A.P, the first term is 25, n^{th} term is 17 and the sum of n terms is 132. Find n and the common difference.
If the 8^{th }term of an A.P is 37 and the 15^{th} term is 15 more than the 12^{th} term, find the A.P. Also, find the sum of first 20 terms of A.P.
Find the sum of all multiples of 7 lying between 300 and 700.
The sum of n natural numbers is 5n^{2} + 4n. Find its 8^{th} term
The fourth term of an A.P. is 11 and the eighth term exceeds twice the fourth term by 5. Find the A.P and the sum of first 50 terms
Chapter 10  Arithmetic Progression Excercise Ex. 10(D)
Find three numbers in A.P. whose sum is 24 and whose product is 440.
The sum of three consecutive terms of an A.P. is 21 and the sum of their squares is 165. Find these terms
The angles of a quadrilateral are in A.P. with common difference 20^{o}. Find its angles.
Divide 96 into four parts which are in A.P and the ratio between product of their means to product of their extremes is 15 :7.
Find five numbers in A.P. whose sum is 12.5 and the ratio of the first to the last terms is 2: 3.
We know that,
Sum of n terms of an A.P =
Let the first term be 2x and the last term be 3x.
∴ Sum of 5 terms of an A.P =
First term = 2x =2 × 1 = 2 and the last term = 3x = 3 × 1 = 3
n^{th} term of an A.P. is given by
t_{n} = a + (n  1)d
⇒ a_{5} = 2 + (5  1)d
⇒ 3 = 2 + 4d
⇒ 1 = 4d
⇒ d = = 0.25
Therefore the five numbers in an A.P are 2, 2.25, 2.50, 2.75 and 3.
Split 207 into three parts such that these parts are in A.P. and the product of the two smaller parts is 4623.
The sum of three numbers in A.P. is 15 and the sum of the squares of the extreme terms is 58. Find the numbers.
Find four numbers in A.P. whose sum is 20 and the sum of whose squares is 120.
Insert one arithmetic mean between 3 and 13
The angles of a polygon are in A.P with common difference 5^{o}. If the smallest angle is 120^{o}, find the number of sides of the polygon
Let the number sides be n

Chapter 10  Arithmetic Progression Excercise Ex. 10(E)
Two cars start together in the same direction from the same place. The first cargoes at uniform speed of 10 km h^{1}. The second car goes at a speed of 8 km h^{1} in the first hour and thereafter increasing the speed by 0.5 km h^{1} each succeeding hour. After how many hours will the two cars meet?
Let the two cars meet after n hours, then

A sum of Rs. 700 is to be paid to give seven cash prizes to the students of a school for their overall academic performance. If the cost of each prize is Rs. 20 less than its preceding prize; find the value of each of the prizes.
An article can be bought by paying Rs. 28,000 at once or by making 12 monthly installments. If the first installment paid is Rs. 3,000 and every other installments is Rs. 100 less than the previous one, find :
(i) amount of installments paid in the 9^{th} month
(ii) total amount paid in the installment scheme
A manufacturer of TV sets produces 600 units in the third year and 700 units in the 7^{th} year. Assuming that the production increases uniformly by a fixed number every year find:
(i) the production in the first year.
(ii) the production in the 10^{th} year.
(iii) the total production in 7 years.
Since the production increases uniformly by a fixed number every year, he sequence formed by the production in different years is an A.P.
Let the production in the first year = a
Common difference = Number of units by which the production increases every year = d
Mrs. Gupta repays her total loan of Rs. 1,18,000 by paying installments every month. If the installments for the first month is Rs. 1,000 and it increases by Rs. 100 every month, What amount will she pays as the 30^{th} installments of loan? What amount of loan she still has to pay after the 30^{th} installments
Chapter 10  Arithmetic Progression Excercise Ex. 10(F)
The 6^{th} term of an A.P. is 16 and the 14^{th} term is 32. Determine the 36^{th} term.
Let 'a' be the first term and 'd' be the common difference of the given A.P.
Now, t_{6} = 16 (given)
⇒ a + 5d = 16 ….(i)
And,
t_{14} = 32 (given)
⇒ a + 13d = 32 ….(ii)
Subtracting (i) from (ii), we get
8d = 16
⇒ d = 2
⇒ a + 5(2) = 16
⇒ a = 6
Hence, 36^{th} term = t_{36} = a + 35d = 6 + 35(2) = 76
If the third and the 9^{th} term of an A.P. be 4 and 8 respectively, find which term is zero?
An A.P. consists of 50 terms of which 3^{rd} term is 12 and the last term is 106. Find the 29^{th} term of the A.P.
For a given A.P.,
Number of terms, n = 50
3^{rd} term, t_{3} = 12
⇒ a + 2d = 12 ….(i)
Last term, l = 106
⇒ t_{50} = 106
⇒ a + 49d = 106 ….(ii)
Subtracting (i) from (ii), we get
47d = 94
⇒ d = 2
⇒ a + 2(2) = 12
⇒ a = 8
Hence, t_{29} = a + 28d = 8 + 28(2) = 8 + 56 = 64
Find the arithmetic mean of:
(i) 5 and 41
(ii) 3x  2y and 3x + 2y
(iii) (m + n)^{2 }and (m  n)^{2}
Find the sum of first 10 terms of the A.P.
4 + 6 + 8 + ……
Here,
First term, a = 4
Common difference, d = 6  4 = 2
n = 10
Find the sum of first 20 terms of an A.P. whose first term is 3 and the last term is 57.
Here,
First term, a = 3
Last term, l = 57
n = 20
How many terms of the series 18 + 15 + 12 + …….. when added together will give 45?
Here, we find that
15  18 = 12  15 = 3
Thus, the given series is an A.P. with first term 18 and common difference 3.
Let the number of term to be added be 'n'.
⇒ 90 = n[36  3n + 3]
⇒ 90 = n[39  3n]
⇒ 90 = 3n[13  n]
⇒ 30 = 13n  n^{2}
⇒ n^{2}  13n + 30 = 0
⇒ n^{2}  10n  3n + 30 = 0
⇒ n(n  10)  3(n  10) = 0
⇒ (n  10)(n  3) = 0
⇒ n  10 = 0 or n  3 = 0
⇒ n = 10 or n = 3
Thus, required number of term to be added is 3 or 10.
The n^{th} term of a sequence is 8  5n. Show that the sequence is an A.P.
t_{n} = 8  5n
Replacing n by (n + 1), we get
t_{n+1} = 8  5(n + 1) = 8  5n  5 = 3  5n
Now,
t_{n+1}  t_{n} = (3  5n)  (8  5n) = 5
Since, (t_{n+1}  t_{n}) is independent of n and is therefore a constant.
Hence, the given sequence is an A.P.
Find the general term (n^{th} term) and 23^{rd} term of the sequence 3, 1, 1, 3, ….. .
The given sequence is 1, 1, 3, …..
Now,
1  3 = 1  1 = 3  (1) = 2
Hence, the given sequence is an A.P. with first term a = 3 and common difference d = 2.
The general term (n^{th} term) of an A.P. is given by
t_{n} = a + (n  1)d
= 3 + (n  1)(2)
= 3  2n + 2
= 5  2n
Hence, 23^{rd} term = t_{23} = 5  2(23) = 5  46 = 41
Which term of the sequence 3, 8, 13, ........ is 78?
The given sequence is 3, 8, 13, …..
Now,
8  3 = 13  8 = 5
Hence, the given sequence is an A.P. with first term a = 3 and common difference d = 5.
Let the n^{th} term of the given A.P. be 78.
⇒ 78 = 3 + (n  1)(5)
⇒ 75 = 5n  5
⇒ 5n = 80
⇒ n = 16
Thus, the 16^{th} term of the given sequence is 78.
Is 150 a term of 11, 8, 5, 2, ....... ?
The given sequence is 11, 8, 5, 2, …..
Now,
8  11 = 5  8 = 2  5 = 3
Hence, the given sequence is an A.P. with first term a = 11 and common difference d = 3.
The general term of an A.P. is given by
t_{n} = a + (n  1)d
⇒ 150 = 11 + (n  1)(5)
⇒ 161 = 5n + 5
⇒ 5n = 166
The number of terms cannot be a fraction.
So, clearly, 150 is not a term of the given sequence.
How many two digit numbers are divisible by 3?
How many multiples of 4 lie between 10 and 250?
The sum of the 4^{th} and the 8^{th} terms of an A.P. is 24 and the sum of the sixth term and the tenth is 44. Find the first three terms of the A.P.
The sum of first 14 terms of an A.P. is 1050 and its 14^{th} term is 140. Find the 20^{th} term.
Let 'a' be the first term and 'd' be the common difference of the given A.P.
Given,
S_{14} = 1050
⇒ 7[2a + 13d] = 1050
⇒ 2a + 13d = 150
⇒ a + 6.5d = 75 ….(i)
And, t_{14} = 140
⇒ a + 13d = 140 ….(ii)
Subtracting (i) from (ii), we get
6.5d = 65
⇒ d = 10
⇒ a + 13(10) = 140
⇒ a = 10
Thus, 20^{th} term = t_{20} = 10 + 19d = 10 + 19(10) = 200
The 25^{th} term of an A.P. exceeds its 9^{th} term by 16. Find its common difference.
n^{th} term of an A.P. is given by t_{n}_{}= a + (n  1) d.
⇒ t_{25 } = a + (25  1)d = a + 24d and
t_{9} = a + (9  1)d = a + 8d
According to the condition in the question, we get
t_{25 }= t_{9} + 16
⇒ a + 24d = a + 8d + 16
⇒ 16d = 16
⇒ d = 1
For an A.P., show that:
(m + n)^{th} term + (m  n)^{th} term = 2 × m^{th}^{}term
Let a and d be the first term and common difference respectively.
⇒(m + n)^{th} term = a + (m + n  1)d …. (i) and
(m  n)^{th} term = a + (m  n  1)d …. (ii)
From (i) + (ii), we get
(m + n)^{th} term + (m  n)^{th} term
= a + (m + n  1)d + a + (m  n  1)d
= a + md + nd  d + a + md  nd  d
= 2a + 2md  2d
= 2a + (m  1)2d
= 2[ a + (m  1)d]
= 2 × m^{th}^{}term
Hence proved.
If the n^{th} term of the A.P. 58, 60, 62,.... is equal to the n^{th} term of the A.P. 2, 5, 12, …., find the value of n.
In the first A.P. 58, 60, 62,....
a = 58 and d = 2
t_{n} = a + (n  1)d
⇒ t_{n}_{} = 58 + (n  1)2 …. (i)
In the first A.P. 2, 5, 12, ….
a = 2 and d = 7
t_{n} = a + (n  1)d
⇒ t_{n}_{}= 2 + (n  1)7 …. (ii)
Given that the n^{th} term of first A.P is equal to the n^{th} term of the second A.P.
⇒58 + (n  1)2 = 2 + (n  1)7 … from (i) and (ii)
⇒58 + 2n  2 = 2 + 7n  7
⇒ 65 = 5n
⇒ n = 15
Which term of the A.P. 105, 101, 97 … is the first negative term?
Here a = 105 and d = 101  105 = 4
Let a_{n} be the first negative term.
⇒ a_{n} < 0
⇒ a + (n  1)d < 0
⇒ 105 + (n  1)(4)<0
⇒ 105  4n + 4 <0
⇒ 109  4n < 0
⇒ 109 <4n
⇒ 27.25 < n
The value of n = 28.
Therefore 28^{th} term is the first negative term of the given A.P.
How many three digit numbers are divisible by 7?
The first three digit number which is divisible by 7 is 105 and the last digit which is divisible by 7 is 994.
This is an A.P. in which a = 105, d = 7 and t_{n} = 994.
We know that n^{th} term of A.P is given by
t_{n} = a + (n  1)d.
⇒ 994 = 105 + (n  1)7
⇒ 889 = 7n  7
⇒ 896 = 7n
⇒ n = 128
∴ There are 128 three digit numbers which are divisible by 7.
Divide 216 into three parts which are in A.P. and the product of the two smaller parts is 5040.
Let the three parts of 216 in A.P be (a  d), a, (a + d).
⇒a  d + a + a + d = 216
⇒ 3a = 216
⇒ a = 72
Given that the product of the two smaller parts is 5040.
⇒a(a  d ) = 5040
⇒ 72(72  d) = 5040
⇒ 72  d = 70
⇒ d = 2
∴ a  d = 72  2 = 70, a = 72 and a + d = 72 + 2 = 74
Therefore the three parts of 216 are 70, 72 and 74.
Can 2n^{2}  7 be the n^{th} term of an A.P? Explain.
We have 2n^{2}  7,
Substitute n = 1, 2, 3, … , we get
2(1)^{2}  7, 2(2)^{2}  7, 2(3)^{2}  7, 2(4)^{2}  7, ….
5, 1, 11, ….
Difference between the first and second term = 1  (5) = 6
And Difference between the second and third term = 11  1 = 10
Here, the common difference is not same.
Therefore the n^{th} term of an A.P can't be 2n^{2}  7.
Find the sum of the A.P., 14, 21, 28, …, 168.
Here a = 14 , d = 7 and t_{n} = 168
t_{n} = a + (n  1)d
⇒ 168 = 14 + (n  1)7
⇒ 154 = 7n  7
⇒ 154 = 7n  7
⇒ 161 = 7n
⇒ n = 23
We know that,
Therefore the sum of the A.P., 14, 21, 28, …, 168 is 2093.
The first term of an A.P. is 20 and the sum of its first seven terms is 2100; find the 31^{st} term of this A.P.
Here a = 20 and S_{7} = 2100
We know that,
To find: t_{31} =?
t_{n} = a + (n  1)d
Therefore the 31^{st} term of the given A.P. is 2820.
Find the sum of last 8 terms of the A.P. 12, 10, 8, ……, 58.
First we will reverse the given A.P. as we have to find the sum of last 8 terms of the A.P.
58, …., 8, 10, 12.
Here a = 58 , d = 2
Therefore the sum of last 8 terms of the A.P. 12, 10, 8, ……, 58 is 408.
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