On this page, you will find Arithmetic Progressions Class 10 Notes Maths Chapter 5 Pdf free download. CBSE NCERT Class 10 Maths Notes Chapter 5 Arithmetic Progressions will seemingly help them to revise the important concepts in less time.

## CBSE Class 10 Maths Chapter 5 Notes Arithmetic Progressions

### Arithmetic Progressions Class 10 Notes Understanding the Lesson

We have observed many things in our daily life, follow a certain pattern.

(a) 1, 4, 7, 10, 13, 16, …….

(b) 15, 10, 5, 0, -5, -10,………….

(c) 1,\(\frac{1}{2}\),0,\(-\frac{1}{2}\)………………

These patterns are generally known as sequence. Two such sequences are arithmetic and geometric sequences. Let us investigate the Arithmetic sequence.

1. Sequence: A sequence is a ordered list of numbers.

Terms: The various numbers occurring in a sequence are called its terms. Terms of sequence are denoted by a_{1} a_{2}, a_{3}, …………… a_{n}.

2. Arithmetic Progression: An arithmetic progression is a sequence of numbers such that the difference between the consecutive terms are equal.

3. Common difference: The difference between two consecutive terms of an arithmetic progression is called common difference.

d = a_{2} – a_{1
}d = a_{3} – a_{2} _{
}d = a_{4} – a_{3}

……………..

……………..

d = a_{n }– a_{n-1}

4. Finite Arithmetic Progression: A sequence which has finite or definite number of terms is called finite sequence.

Example, (1, 3, 5, 7, 9)… which has 5 terms.

5. Infinite Arithmetic Progression: A sequence which has indefinite or infinite number of terms is called infinite arithmetic progression.

Example, 1, 2, 3, 4, 5, …

In general, arithmetic progression can be written as a, a + d, a + 2d, where a is the first term and d is called the common difference i.e. difference between two consecutive terms.

6. General form of an AP: Let a be the first term and d is the common difference then the AP is

Here

a_{1} = a (we take) (a is first term of AP)

a_{2} = a_{1} + d = a + d

a_{3} = a_{2} + d = a + d + d = a + 2d

a_{4} = a_{3}+ d = a + 2d + d = a + 3d

…………….

……………

a_{n} = a + (n – 1) d

i.e. AP is a, a + d, a + 2d, a + 3d,………… , a + (n – 1)d.

n^{th} term of AP = a + (n -1)d

Note: Common difference of AP can be positive, negative or zero.

1. n^{th} term or General term of an AP

n^{th} term of an AP = a + (n – 1) d where

a → first term of the AP

n → number of terms

d→common difference of an AP.

2. n^{th} term of an AP from the end: Let us consider an AP where first term a and common difference is If m is number of terms in the AP. then

n^{th} term from the end = [m – n + 1]^{th} term from the beginning.

n^{th} term from the end = a + (m-n +1 – 1)d – a + (m – n) d

It l is the last term of the AP, then n^{th} term from the end is the n^{th} term of an AP where first term is l and common difference is – d

n^{th} term from the end – 1 + (n – 1) (-d)

= 1 – (n – 1) d

Sum of first n terms of an AP

Let S_{n} denote the sum of first n terms of an AP

S_{n} = a + a + d + a + 2d + a + 3d …. + a + (n – 1)d ……….. (1)

Rewriting the terms in reverse order.

S_{n} = a + (n – 1) + a + (n – 2)d + a + (n-3)d + ………….+a ……… (2)

Adding equations (1) and (2)

2S_{n} = [2a + (n – 1)d] + [2a + (n-1)d] + … + [2a + (n – 1)d]

2S_{n} = n[2a + (n – 1)d]

S_{n}=\(\frac{n}{2}\)[2a+(n-1)d]

We can Write

S_{n}=\(\frac{n}{2}\)[a+a+(n-1)d] [l=a+(n-1)d]

S_{n}=\(\frac{n}{2}\)[a+l]

Note:

(i) The Tith term of an AP = S_{n} – S_{n-1} or a_{n} = S_{n+1} – S_{n
}Sum of first n positive integer

\(S_{n}=\frac{n(n+1)}{2}\)

(iii) Sum of n odd positive integer = n^{2
}(iv) Sum of n even positive integer = n(n + 1)

## Leave a Reply