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# Arithmetic Progressions CBSE Notes Class 10 Maths Chapter 5 PDF

## Arithmetic Progression

An **arithmetic progression** is a list of numbers in which each term is obtained by adding a fixed number to the preceding term except the first term. This fixed number is called the common difference of the AP and the common difference of AP can be positive, negative or zero.

If the sequence t_{1}, t_{2}, t_{3}, …, t_{n-1}, t_{n} are in AP, then t_{2} – t_{1} = t_{3} – t_{2} = t_{4} – t_{3} = … = t_{n} – t_{n-1 }= d (common difference).

The sequence a, a + d, a + 2d, a + 3d, . . . represents an arithmetic progression where a is the first term and d the common difference. This is called the general form of an AP.

### Class 10 Maths Chapter 5 Examples 1:

Write first four terms of the AP, when the first term a = 10 and the common difference d = 10.

The arithmetic progression is

a, a + d, a + 2d, a + 3d, . . . ⇒ 10, 20, 30, 40, …

### Class 10 Maths Chapter 5 Examples 2:

Check whether the sequence 2, 4, 8, 16, . . . is in AP or not.

4 – 2 ≠ 8 – 4. Thus, the given sequence is NOT in AP.

### Class 10 Maths Chapter 5 Examples 3:

For the AP, 0.6, 1.7, 2.8, 3.9, . . ., write the first term and the common difference.

The first term of the AP is a = 0.6 and the common difference is d = 1.7 – 0.6 = 1.1.

## n^{th} Term of General Term of an AP

If the sequence t_{1}, t_{2}, t_{3}, …, t_{n-1}, t_{n} are in AP, then the general term of the AP is **t _{n} = a + (n – 1)d**, where

**a**is the first term (

**a = t**) and

_{1}**d**is the common difference (

**d = t**= t

_{2}– t_{1}_{3}– t

_{2}= … etc).

## n^{th} Term of General Term of an AP from Last

Let **a** be the first term (**a = t _{1}**) and

**d**be the common difference (

**d = t**= t

_{2}– t_{1}_{3}– t

_{2}= … etc) of an AP having total N terms. Then, the

**n**from the end or last of the AP is the

^{th}term**(N – n + 1)**term from the beginning of the AP.

^{th}### Class 10 Maths Chapter 5 Examples 4:

Find the 10th term of the AP : 2, 7, 12, . . .

Given that a = 2, d = 7 – 2 = 5.

Since t_{n} = a + (n – 1)d

⇒ t_{10} = 2 + (10 – 1)(5) = 2 + 9 × 5 = 2 + 45 = 47

### Class 10 Maths Chapter 5 Examples 5:

Check whether 301 is a term of the list of numbers 5, 11, 17, 23, . . .

t_{n} = a + (n – 1)d

⇒ 301 = 5 + (n – 1)6

⇒ 301 = 5 + 6n – 6

⇒ 301 = 6n – 1

⇒ 6n = 302

⇒ n = 151/3 (Not Natural number).

Thus, 301 is not a term of the given sequence.

### Class 10 Maths Chapter 5 Examples 6:

Which term of the AP : 21, 18, 15, . . . is – 81?

Given that a = 21, d = 18 – 21 = -3.

Since t_{n} = a + (n – 1)d

⇒ –81 = 21 + (n – 1)(-3)

⇒ –81 = 21 – 3n + 3

⇒ 3n = 24 + 81 = 105

⇒ 3n = 105

⇒ n = 105 / 3 = 35

## Sum of n Terms of an AP

Let **a** be the first term (**a = t _{1}**) and

**d**be the common difference (

**d = t**= t

_{2}– t_{1}_{3}– t

_{2}= … etc) of an AP having total n terms. Then, the

**sum of n terms of AP**is

If the last term or n^{th} term of the AP is known, then the **sum of n terms of AP** is

The sum of first n natural numbers is S = 1 + 2 + 3 + … + n = n(n + 1)/2.

### Class 10 Maths Chapter 5 Examples 7:

Find the sum of the first 22 terms of the AP : 8, 3, –2, . . ..

Given that a = 8, d = 3 – 8 = –5, n = 22.

### Class 10 Maths Chapter 5 Examples 8:

Find the sum of the first 1000 positive integers.

Required sum S = 1 + 2 + 3 + … + 1000 = 1000(1000 + 1)/2 = 500 × 1001 = 500500.

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