# Introduction to Arithmetic Progressions | Class 10 Maths

A progression is a sequence or series of numbers in which they are arranged in a particular order such that the relation between the consecutive terms of series or sequence is always constant. In a progression, it is possible to obtain the n_{th }term of the series.

In mathematics, there are 3 types of progressions:

- Arithmetic Progression(AP)
- Geometric Progression(GP)
- Harmonic Progression(HP)

**Arithmetic Progression (AP)** also known as Arithmetic Sequence is a sequence or series of numbers such that the common difference between two consecutive numbers in the series is constant.

**For example:**

**Series 1: 1,3,5,7,9,11….**

In this series, the common difference between any two consecutive numbers is always 2.

**Series 2: 28,25,22,19,16,13….**

In this series, the common difference between any two consecutive numbers is strictly -3.

**Terminology and Representation**

- Common difference, d = a
_{2}– a_{1}= a_{3}– a_{2}= ……. = a_{n}– a_{n – 1} - a
_{n }= n_{th}term of Arithmetic Progression - S
_{n }= Sum of first n elements in the series

### General Form of an AP

If **a** is taken as the first term and** d** is taken as the common difference, then the N^{th} term of the AP will be given by the formula:

So by computing the n terms of an AP with the upgiven formula, the general form of the AP is as follows:

**Example: Find the 35 ^{th} term of the series 5,11,17,23…..**

**Solution:**

In the given series,

a = 5, d = a

_{2 }– a_{1 }= 11 – 5 = 6, n = 35We have to find out the 35

^{th}_{ }term, hence, apply the formulae,a

_{n }= a + (n – 1)da

_{n }= 5 + (35 – 1) x 6a

_{n }= 5 + 34 x 6a

_{n }= 209Hence 209 is the 35

_{th}term.

### Sum of n Terms of Arithmetic Progression

The formula for the arithmetic progression sum is,

**S _{n} = (n/2)[2a + (n − 1) × d]**

### Derivation of the Formula

Let ‘l’ denote the n_{th} term of the series and S_{n} be the sun of first n terms of AP a, (a+d), (a+2d), …., a+(n-1)d then,

S

_{n }= a_{1 }+ a_{2 }+ a_{3 }+ ….a_{n-1 }+ a_{n}S

_{n }= a + (a + d) + (a + 2d) + …….. + (l – 2d) + (l – d) + l …(1)Writing the series in reverse order, we get,

S

_{n }= l + (l – d) + (l – 2d) + …….. + (a + 2d) + (a + d) + a …(2)Adding equation (1) and (2),

2S

_{n }= (a + l) + (a + l) + (a + l) + …….. + (a + l) + (a + l) + (a + l)2S

_{n }= n(a + l)S

_{n }= (n/2)(a + l) …(3)Hence, the formulae for finding the sum of a series is,

S_{n }= (n/2)(a + l)where,

a is the first term

l is the last term of the series and

n is the number of terms in the series

Replacing the last term l by the n

^{th}term in equation 3 we get,n

^{th}term = a + (n – 1)dS

_{n }= (n/2)(a + a + (n – 1)d)

S_{n }= (n/2)(2a + (n – 1) x d)

Note: The consecutive terms in an Arithmetic Progression can also be represented as,

…….., a-3d , a-2d, a-d, a, a+d, a+2d, a+3d, ……..

### Sample Problems on Arithmetic Progressions

**Problem 1. Find the sum of the first 35 terms of series 5,11,17,23…..**

**Solution:**

In the given series,

a = 5, d = a

_{2 }– a_{1 }= 11 – 5 = 6, n = 35S

_{n }= (n/2)(2a + (n – 1) x d)S

_{n }= (35/2)(2 x 5 + (35 – 1) x 6)S

_{n }= (35/2)(10 + 34 x 6)S

_{n }= (35/2)(10 + 204)S

_{n }= 35 x 214/2S

_{n }= 3745

**Problem 2. Find the sum of the series when the first term of the series is 5 and the last of the series is 209 and the number of terms in the series is 35.**

**Solution:**

In the given series,

a = 5, l = 209, n = 35

S

_{n }= (n/2)(a + l)S

_{n }= (35/2)(5 + 209)S

_{n }= 35 x 214/2S

_{n }= 3745

**Problem 3. 21 Rupees is divided among three brothers where the three parts of money are in AP and the sum of their squares is 155. Find the largest amount.**

**Solution:**

Let the three parts of money be (a-d), a, (a+d), as the distributed amount is in AP

Given that

(a – d) + a + (a + d) = 21

Therefore,

3a = 21

a = 7

Again, (a – d)

^{2}+ a^{2}+ (a + d)^{2}= 155a

^{2}+ d^{2}– 2ad + a^{2}+ a^{2}+ d^{2}+ 2ad = 1553a

^{2}+ 2d^{2}= 155Putting the value of ‘a’ we get,

3(7)

^{2}+ 2d^{2}= 1552d

^{2}= 155 – 147d

^{2}= 4d = ±2

The three parts of distributed money are:

a + d = 7 + 2 = 9

a = 7

a – d = 7 – 2 = 5

Hence, the largest part is Rupees 9