2. PREFACE
The National Curriculam Frame work, 2005 suggests the need for developing the ability
for mathematisation in the students. It points out that the aim of learning mathematics is not
merely being able to do quantitative calculations also to develop abilities in the students that
would enable her/his to redefine her/his relationship with the world. It recommends that
mathematics needs to slowly move towards abstraction even through it starts from concrete
experiences and models. The ability to generalize and perceive patterns in an important step in
being able to relate to the abstract and logic governed nature of the subject.
I hope that this textbook would help all students in their attempt to learn mathematics and
would build in them the ability to appreciate its power and beauty. And I hope to strengthen the
Foundation of Mathematics on which further engagement with studies as well as her daily life
would become possible in an enriched manner.

3. CONTENTS
PREFACE
1. ARITHMETIC SEQUENCE
1.1 Introduction
1.2 Definition of Arithmetic Progression
1.3 nth term of an Arithmetic Sequence
1.4 Sum to n terms of an AP
1.5 Application of AP

4. ‘
1. ARITHMETIC SEQUENCE
1.1 Introductions
We have learnt about different types of numbers in our previous classes. When we say
that a collection of objects is listed in a sequence or progression, we usually mean that the
collection is ordered in such a way that it has an identified first number, second number, third
number and so on. For example, the amount of money deposited in a bank, over a number of
years form a sequence or progression. Sequences have important applications in several spheres
of human activities.

5. 1.2 Definition of Arithmetic Progression
Arithmetic Progression (AP) is an arrangement of numbers in which terms increase or
decrease regularly by the same constant. The constant is called the common difference of the
Arithmetic Sequence or Arithmetic Progression (AP). We denote the first term of the AP by a,
the common difference by d, and the last term by l. Consider the following sequence.
4, 7, 10, 13, …………..
In this sequence, the difference of any consecutive terms is 3. Therefore this sequence is an AP.
Here 3 is the common difference
i.e d = 3
obviously, the first terms is 4 i.e. a = 4
Exercise 1.2
Check whether the following sequences are arithmetic sequence or not. If which are arithmetic
sequences, write their first term and common difference.
(i) 1, 6, 11, 17, ………….
(ii) -1, -3, -5. -7, …………
(iii) 1/4, 1/5, 1/6, 1/7, ……..
(iv)
1
4
,
1
5
,
1
6
,
1
7
………
(v) 8, 9, 11. 17, …….

6. 1.3 nth term of an arithmetic sequence
Consider an arithmetic sequence of first term ‘a’ and the common difference ‘d’.
i.e. AP is,
a, a+d, a+2d, ……….
Let a1, a2, a3, ……….. denote its first term, second term, third term, and so on ..
Then
a1 = a
a2 = a + d
a3 = a + 2d
………………………………
………………………………
Proceeding like this,
We get the nth term
an = a + (n – 1) d

7. Example 1 : Consider the sequence 4, 6, 8, 10, …………
Find the 25th term of this sequence?.
Solution :
Given sequence is
4, 6, 8, 10, ……….
Obviously, this sequence is an AP
Here a = 4, d = 6 – 4 = 2
We know that an = a + (n – 1) d
a25 = 4 + (25 – 1) x 2 = 52
Example 2 : Is 244 a term of the sequence
1, 7, 13, 19, ……
Solution :
Given sequence is 1, 7, 13, 19, …… It is an AP with a = 1, d = 6
Suppose that 244 belongs to this AP.
Let an = 244
Then 244 = 1 + (n – 1) x 6
Solving we get n =
249
6
, which is not a natural n number
Hence 244 is not a term of this sequence.

8. Exercise 1.3
I. (i) In an AP, the first term is -4 and the common difference is 5. Find its 36th term ?
(ii) In an AP, a = 1. an = 45, n = 96. Find ‘d’.
II. (i) Is 395 a term of the AP 5, 10, 15, ……………. ?
(ii) Find the 75th term of the AP 1,
1
2
,
3
2
, …….
1.4 Sum to n terms of an AP
Let us consider an AP with first term ‘a’ and common difference d
i.e. a, a + d, a + 2d, ……
If Sn denote the sum of the first n norms of the AP, then we write
Sn = a + (a + d) + (a + 2d) + .. + a + (n-1) d
Also, write the above sum is reverse order, then
Sn = a + (n-1) d + a + (n – 2) d + … + (a + d) + a
Adding the above equations, we get Sn+Sn = [2a+(n-1)d] + [2a+(n-1)d] + ….. n times
i.e 2Sn = n [2a + (n-1) d]
Sn =
푛
2
[2a + (n – 1)d]

9. Let us consider some examples.
Example 1 : Find the sum of the first 22 terms of the AP 8, 3, -2, ……..
Solution : Here a = 8, d = 3 - 8 = -5, n = 22
We know that S =
푛
2
[2a + (n -1) d]
S =
22
2
[16 + 21 (- 5)]
= 11 ( 16 – 105 )
= 11 (- 89)
= - 979
So, the sum of the first 22 terms of the AP is – 979.
Example 2 : If the sum of the first 14 terms of an AP is 1050 and its first terms is 10. Find the
20th term ?
Solution : Here S14 = 1050, n = 14, a = 10
Sn =
푛
2
[2a + (n-1) d]
1050 =
14
2
[20 + 13d]
= 140 + 91d
91d = 910
D =
910
91
= 10
Therefore, d20 = 10 + (20 – 1) x 10 = 200

10. i.e. 20th term is 200
Example 3 : How many terms of the AP 24, 21, 18, ……. must taken so that their sum is 78 ?
Solution : Here a = 24, d = -3, Sn = 78. We need to find n.
We know that Sn =
푛
2
[2a + (n-1) d]
So, 78 =
푛
2
[51 + 3n]
Or 3n2 – 51n + 156 = 0
Or n2 – 17n + 52 = 0
(n – 4) (n – 13) = 0
N = 4 or 13
Exercise 1.4
I. Find the sum of the following APs.
(i) 2, 7, 12, …………. to 10 terms
(ii) -37, -33, -29, ……. to 12 terms
(iii)
1
15
, 1
12
, 1
10
,……. to 11 terms
II. (i) How many terms of the AP. 9, 17, 25, …… must be taken to give a sum of 636 ?.
(ii) Find the sum 34 + 32 + 30 + ….. + 10

11. 1.5 Application of Arithmetic Sequence
Arithmetic sequence have important applications in several spheres of human activities.
This section recognizes some problems connected with various fields in human activities and
find out their solutions using the theory of arithmetic progressions.
Let us consider the following problems.
Example 1 : A manufacturer of TV sets produced 600 sets in the third year and 700 sets in the
seventh 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 10th year
(iii) the total production in first 7 years.
Solution : (i) Since the production increases uniformly by a fixed number every year, the
number of TV sets manufactured in 1st , 2nd , 3rd , ….. years will form an AP. Let us denote the
number of TV sets manufactured in nth year by an.
Then a3 = 600 and a7 = 700
Or a+ 2d = 600, a + 6d = 700
→ a = 550, d = 25
Therefore, production of TV sets in the first year is 550.

12. (ii) Now a10 = a + 9d
= 550 + 9 x 25
= 775
So, the production of TV sets in the 10th year is 775.
(iii) Also, S7 =
7
2
, [ 2 x 550 = (7 – 1) x25] = 4375
Thus, the total production of TV sets in first 7 years is 4375.
Example 2 : In a flower bed, there are 23 rose plants in the first row, 21 in the second, 19 in the
third, and so on. There are 5 rose plants in the last row. How many rows are there in the flower
bed ?
Solution : The number of rose plants in the first row, 2nd row, 3rd row …….. are 23. 21. 19, ….5.
It forms an AP. Let the number of rows in the flower bed be n.
Then a = 23, d = 21 – 23 = -2
an = 5
As, an = a + (n – 1) d
5 = 23 + (n – 1) (-2)
Solving we get n = 10
So there are 10 roses in the flower bed.

13. Example 3 : The income of a person is Rs. 3,00,000/- in the first year and he receives an
increase of Rs. 10,000/- to his income per year for the next 19 years. Find the total
amount he receives in 20 years ?
Solution : Here we have an AP with a = 3,00,000, d = 10,000, n = 20, using the formula, we get
S20 =
20
2
, [ 600000 + 19 x 10000] = 10 (790000)
= 79,00,000
Hence the person received Rs. 79,00,000 as the total amount at the end of 20 years.
Exercise (1.5)
1) A man starts repaying a loan as first installment of Rs. 100/-. If he increases the
installment by Rs. 5/- every month, what amount he will pay in the 30th installment ?
2) Manu saved Rs. 5 in the first week of a year and then increased his weekly savings by Rs.
1.75. If in the nth week, his weekly savings become Rs. 20.75. Find n.

14. Answers
Exercise 1.2
(i) AP : a = 1, d = 5
(ii) AP : a = -1, d = -2
(iii) Not AP
(iv) Not AP
Exercise 1.3
I (i) 171
(ii) d =
44
95
II (i) Yes
(ii)
175
3
Exercise 1.4
I (i) 245
(ii) - 180
(iii)
33
20
II (i) 44
(ii) 440
Exercise 1.5
(i) 245
(ii) 10