# SCERT Kerala Maths Class 10/ Arithmetic Sequences

SCERT Kerala Maths Class 10/ Arithmetic sequence is about the textbook solutions of Chapter 1 Arithmetic Sequences. Here you can find out practice problems for Class 10 Mathematics.

SCERT Kerala Maths Class 10 / Arithmetic Sequences/Textbook solutions.
Kerala State Board SSLC Mathematics/ Chapter 1
Position and Term:

1. In each of the arithmetic sequences below, some terms are missing and their positions are marked with —. Find them.
i) 24, 42, —, —
ii) —, 24, 42, —
iii) —, —, 24, 42, ————
iv) 24, —, 42, —,
v) —, 24, —, 42, ———-
vi) 24, —, —, 42, ——-
Solution:
i) Let the third number be x and fourth number be y.
Then the sequence will be 24, 42, x, y —————
So 42 – 24 = x – 42
x = 84 – 24 = 60
Also 60 – 42 = y – 60
y = 120 – 42 = 78
Required arithmetic sequence is 24, 42, 60, 78 —————-
ii) Let the first number be x and fourth number be y.
Then the sequence will be x, 24, 42, y ————-
So 24 – x = 42 – 24
24 – x = 18
x = 24 – 18 = 6
Also 42 – 24 = y – 42
y = 84 – 24 = 60
Required arithmetic sequence is 6, 24, 42, 60 ——————
iii) Let the first number be x and second number be y
Then the sequence will be x, y, 24, 42, ————-
So y – x = 24 – y
2y – x = 24
Also 24 – y = 42 – 24
24 – y = 18
y = 24 – 18 = 6
Substitute the value of y in the equation 2y – x = 24, we get 12 – x = 24
Therefore, x = 12 – 24 = -12
Required arithmetic sequence is -12, 6, 24, 42 —————-
iv) Let second term be x and fourth term be y
Then the sequence will be 24, x, 42, y, ————-
So x – 24 = 42 –x
2x = 42 + 24
2x = 66
x = 66/2 = 33
Also 42 – 33 = y – 42
y = 9 + 42 = 51
Required arithmetic sequence is 24, 33, 42, 51, ————–
v) Let first term be x and third term be y
Then the sequence will be x, 24, y, 42, ——-
So 24 – x = y – 24
Also y – 24 = 42 – y
2y = 42 + 24 = 66
y = 66/2 = 33
Substitute the value of y in 24 – x = y – 24
24 – x = 33 – 24
x = 48 – 33 = 15
Required arithmetic sequence is 15, 24, 33, 42, ————-
vi) Let the second term be x and third term be y
Then the sequence will be 24, x, y, 42, ————
So x – 24 = y –x, which implies 2x – y = 24 ————– (1)
Also y – x = 42 – y, which implies 2y – x = 42 ——— (2)
Solve the equations (1) and (2) we get
Multiply equation (1) with 2, 4x – 2y = 48
Second equation as –x + 2y = 42
Add these two equations, 3x = 90
x = 90/3 = 30
Substitute the value of x in 2x – y = 24
60 – y = 24
y = 60 – 24 = 36
Required arithmetic sequence is 24, 30, 36, 42, —————–
2. The terms in two positions of some arithmetic sequences are given below. Write the first five terms of each:
i) 3rd term 34, 6th term 67
ii) 3rd term 43, 6th term 76
iii) 3rd term 2, 5th term 3
iv) 4th term 2, 7th term 3
v) 2nd term 5, 5th term 2
Solution:
i) To get the 6th term from the 3rd term, we must add the common difference 6 – 3 = 3 times.
Actual number to be added is 67 – 34 = 33
Thus 3 times the common difference is 33 and so the common difference is 33/3 = 11
Second term = 34 – 11 = 23
First term = 23 – 11 = 12
First 5 terms are 12, 23, 34, 45, 56
ii) To get the 6th term from the 3rd term, we must add the common difference 6 – 3 = 3 times.
Actual number to be added is 76 – 43 = 33
Thus thrice the common difference = 33 and so the common difference is 33/3 = 11
Second term = 43 – 11 = 32
First term = 32 – 11 = 21
First 5 terms are 21, 32, 43, 54, 65
iii) To get the 5th term from the 3rd term, we must add the common difference 5 – 3 = 2 times.
Actual number to be added is 3 – 2 = 1
Thus twice the common difference = 1
Common difference = ½
Second term = 2 – ½ = 3/2
First term = 3/2 – ½ = 2/2 = 1
First 5 terms are 1, 3/2, 2, 5/2, 3
iv) To get the 7th term from the 4th term, we must add the common difference 7 – 4 = 3 times.
Actual number to be added is 3 – 2 = 1
Thus thrice the common difference = 1
Common difference = 1/3
Third term = 2 – 1/3 = 5/3
Second term = 5/3 – 1/3 = 4/3
First term = 4/3 – 1/3 = 3/3 = 1
First five terms are 1, 4/3, 5/3, 2, 7/3.
v) To get 5th term from the 2nd term, we must add the common difference 5 – 2 = 3 times.
Actual number to be added is 2 – 5 = -3
Thus thrice the common difference = -3
Common difference = -3/3 = -1
First term = 5 – (-1) = 6
First 5 terms are 6, 5, 4, 3, 2
3. The 5th term of an arithmetic sequence is 38 and the 9th term is 66. What is its 25th term?
Solution:
Term difference = 66 – 38 = 28
Position difference = 9 – 5 = 4
Common difference = 28/4 = 7
25th term = 5th term + 20 x common difference
= 38 + 20 x 7
= 38 + 140
= 178
4. Is 101 a term of the arithmetic sequence 13, 24, 35, ——-? What about 1001?
Solution:
Common difference = 24 – 13 = 11
Difference between 101 and 13 is 101 – 13 = 88 = 8 x 11
i.e., 101 = 13 + 8 x 11
Thus 101 is obtained by adding 8 times the common difference with 13.
So 101 is the 9th term.
Difference between 1001 and 13 is 1001 – 13 = 988
988 cannot be written as a multiple of 11.
So 1001 is not a term of the arithmetic sequence.
5. How many three-digit numbers are there, which leave a remainder 3 on division by 7?
Solution:
Three digit numbers which leave a remainder 3 on division by 7 are 101, 108, ———-, 997
Here first term = 101
Common difference = 7 and last term = 997
Term difference = 997 – 101 = 896 = 128 x 7
Or we can write 997 = 101 + 128 x 7
i.e. 997 is the 129th term.
Therefore, 129 three digit numbers are there, which leave a remainder 3 on division by 7.
6. Fill up the empty cells of the given squares such that the numbers in each row and column form arithmetic sequences:
1 — —– 4

7 — —– 28
What if we use some other numbers instead of 1, 4, 28 and 7?
Solution:|
Given 1st term = 1 and 4th term = 4
To get the 4th term from the 1st term, we must add the common difference 4 – 1 = 3 times
Thrice the common difference = 4 – 1 = 3
Common difference = 3/3 = 1
Arithmetic sequence is 1, 2, 3, 4
In the first column, 1st term = 1 and 4th term = 7
Term difference = 7 – 1 = 6
Position difference = 4 – 1 = 3
Common difference = 6/3 = 2
Arithmetic sequence is 1, 3, 5, 7
In the fourth raw, 1st term = 7 and 4th term = 28
Term difference = 28 – 7 = 21
Position difference = 4 – 1 = 3
Common difference = 21/3 = 7
Arithmetic sequence is 7, 14, 21, 28
In the second column, 1st term = 2 and 4th term = 14
Term difference = 14 – 2 = 12
Position difference = 4 – 1 = 3
Common difference = 12/3 = 4
Arithmetic sequence is 2, 6, 10, 14
In the third column, 1st term = 3 and 4th term = 21
Term difference = 21 – 3 = 18
Position difference = 4 – 1 = 3
Common difference = 18/3 = 6
Arithmetic sequence is 3, 9, 15, 21
In the 4th column, 1st term = 4 and 4th term = 28
Term difference = 28 – 4 = 24
Position difference = 4 – 1 = 3
Common difference = 24/3 = 8
Arithmetic sequence is 4, 12, 20, 28
1 2 3 4
3 6 9 12
5 10 15 20
7 14 21 28

1. In the table below, some arithmetic sequences are given with two numbers against each. Check whether each belongs to the sequence or not.
Sequence Numbers Yes/No
11, 22, 33, ——— 123
132
12,23, 34, ——— 100
1000
21, 32, 43, ———- 100
1000
¼, ½, ¾ ,——- 3
4
¾,1 ½, 2 ¼ , —— 3
4
Solution:
i) Given sequence is 11, 22, 33, ——-
Common difference = 22 – 11 = 11
Difference between 123 and 11 is 123 – 11 = 112
We can’t write 112 as a multiple of 11, so 123 is not a term of the arithmetic sequence.
Difference between 132 and 11 is 132 – 11 = 121
We can write 121 as a multiple of 11. So 132 is a term of the arithmetic sequence.
ii) Given sequence is 12, 23, 34, ———-
Common difference = 23 – 12 = 11
Difference between 100 and 12 is 100 – 12 = 88
We can write 88 as a multiple of 11, so 100 is a term of the arithmetic sequence.
Difference between 1000 and 12 is 1000 – 12 = 988
We can’t write 988 as a multiple of 11, so 1000 is not a term of the arithmetic sequence.
iii) Given sequence is 21, 32, 43, ————
Common difference = 32 – 21 = 11
Difference between 100 and 21 is 100 – 21 = 79
We can’t write 79 as a multiple of 11, so 100 is not a term of the arithmetic sequence.
Difference between 1000 and 21 is 1000 – 21 = 979
We can write 979 as a multiple of 11, so 1000 is a term of the arithmetic sequence.
iv) Given sequence is ¼, ½, ¾, —————-
Common difference = ½ – ¼ = ¼
Difference between 3 and ¼ is 3 – ¼ = 11/4
11/4 is a multiple of ¼, so 3 is a term of the given sequence.
Difference between 4 and ¼ is 4 – ¼ = 15/4
15/4 is a multiple of ¼, so 4 is a term of the given sequence.
v) Given sequence is ¾, 1 ½, 2 ¼, ————–
Common difference = 1 ½ – ¾ = 3/2 – ¾ = ¾
Difference between 3 and ¾ is 3 – ¾ = 9/4
9/4 is a multiple of ¾, so 3 is a term of the given sequence.
Difference between 4 and ¾ is 4 – ¾ = 13/4
13/4 is not a multiple of ¾, so 4 is not a term of the given sequence.