# HBSE 10th Class Maths Notes Chapter 5 Arithmetic Progressions

Haryana State Board HBSE 10th Class Maths Notes Chapter 5 Arithmetic Progressions Notes.

## Haryana Board 10th Class Maths Notes Chapter 5 Arithmetic Progressions

Introduction
We must have observed that in nature, many things follow a certain pattern, such as the petals of a sunflower, the grains on a maize cob, the holes of a honeycomb, the spirals on a pineapple etc.

Many times we come across certain specific patterns of numbers, e.g.
3, 6, 9, 12, 15, 18, ……..
9, 16, 25, 36, 49, ……..
2, 4, 6, 8, 10, 12, ……..
– 60, – 50, – 40, – 30, – 20, …….. etc.
These patterns are generally known as sequences. Hence, a sequence may be defined as an arrangement of numbers in some definite order and according to some rule.
The various numbers appearing in a sequence are called its terms. For example, in the first sequence 3 is the first term, it is denoted by a1, 6 is the second term, it is denoted by a2, 9 is the third term, it is denoted by a, and so on.
Generally, we denoted the terms of a sequence by a1, a2, a3, a4, …… etc. or x1, x2, x3, x4, …….. etc.

In general, the number at the place is called the nth term of the sequences and it is denoted by an is also called the general term of the sequence. In this chapter, we shall study special type of sequence in which succeeding terms are obtained by adding, a field number to the preceding terms. Arithmetic Progressions
Consider the following patterns :
1. 1, 2, 3, 4, 5, …..
2. 10, 20, 30, 40, 50, ….
3. 90. 80. 70. 60. 50 ……
4. -10.5, -11, -11.5, -12, -12.5, …
Each of the numbers in the pattern is called a term.
We observe that:
In Pattern (1) : each term is 1 more than the term preceding it.
In Pattern (2) : each term is 10 more than the term preceding it.
In Pattern (3) : each term is 10 less than the term preceding it.
In Pattern (4) : each term is 0.5 less than the term preceding it.
So, an arithmetic progression is a list of numbers in which each term is obtained by adding or subtracting a fixed number to the preceding term except the first term.
The fixed number is called the common difference of the AP. It is denoted by d. It can be positive, negative or zero.
So, d = a2 – a1 = a3 – a2 = ….. = an – an-1
Therefore, the AP with first term a and common difference d is:
a, a + d, a + 2d, a + 3d, a + 4d, …
In other words, sequence a1, a2, a3, …… an… is called an arithmetic progression is the difference of a term and preceding term is always constant. This constant is called the common difference of the AP.
Finite AP : An AP containing finite number of terms is called finite AP.
Infinite AP : An AP containing infinite number of terms is called infinite AP.
If a, b, c are in AP, then $$\frac{a+c}{2}$$ and b is called arithmetic mean of a and c.

nth Term of an AP
nth term of an AP from the start
Let a1, a2, a3, a4, …… be an AP whose first term is a1 and the common difference is d.
Then a1 = a = a + (1 – 1)d ……(1)
a2 = a + d = a + (2 – 1)d … (2)
a3 = a2 + d
= (a + d) + d
= (a + 2d)
= a + (3 – 1)d …..(3)
a4 = a3 + d
= (a + 2d) + d
= a + 3d
= a + (4 – 1)d ……(4)
Observing the pattern in equations (1), (2), (3), (4), we find that : an = a + (n – 1)d
So, the nth term an of the AP with first terma and common difference is given by :
an = a + (n – 1)d
If there are m terms in the AP then am represents the last term which is also denoted by l.
nth Term of an AP from the end.
(1) If there are m terms in an AP whose first term is a and common difference is d, then
nth term from the end = a + (m – n)d
(2) If l is the last term of the AP then, nth term from the end is the nth term of an AP whose first term is l and common difference is – d.
nth term from the end = last term + (n – 1) (n – d)
= l – (n – 1)d Sum of first n terms of an AP
Let a1, a2, a3, …… be an AP whose first term a and common difference is d. Then
a1 = a, a2 = a + d, a3 = a + 2d, a4 = a + 3d, …… an = a + (n – 1)d.
Now, Sn = a1 + a2 + a3 + a4 + … + an-1 + an
⇒ Sn = a + (a + d) + (a + 2d) + (a + 3d) + …… [a + (n – 2)d] + [a + (n – 1)d] … (1)
Writing the above series in reverse order, we get
Sn = a + [a + (n – 1)d] + [a + (n – 2)d] + …… + (a + d) + a …(2)
Adding the corresponding terms of equation (1), and (2), we get
⇒ 2Sn = a + [a + (n – 1)d] + (a + d) + [a + (n – 2d] + … + [a + (n – 1) d] + a
⇒ 2Sn = [2a + (n – 1)d] + [2a + (n – 1)d] + … + [2a + (n – 1)d]
∵ 2a + (n – 1)d repeats n times.
∴ 2Sn = n[2a + (n – 1)d]
⇒ Sn = $$\frac{n}{2}$$[2a + (n – 1)d]
Now, if l is the last term nth term = nth term = a + (n – 1)d.
Then
Sn = $$\frac{n}{2}$$[a + a + (n − 1)d]
Sn = $$\frac{n}{2}$$[a + l]
Hence, the sum of the first n terms of AP is given by :
Sn = $$\frac{n}{2}$$[2a + (n – 1)d]
or Sn = $$\frac{n}{2}$$[a + l]
where l = last term.
Important Results:
(i) Three numbers in AP considered as (a – d), a, (a + d).
(ii) Four numbers in AP considered as (a – 3d), (a – d), (a + d), (a + 3d).
(iii) Five numbers in AP considered as (a – 2d), (a – d), a, (a + d), (a + 2d).