# 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 1Position and Term:Answer the following:**

- 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, —————– - 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 - 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 - 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. - 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. - 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

- 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.