By Truemaths Tech, Last Updated 21 Jul, 2023 6 min read

**Arithmetic Progression Class 10**

This is another important concept * Arithmetic Progression for class 10* students preparing for

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According to the definition given in the **NCERT book**, “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 it can be positive, negative or zero”.

There are some examples for better understanding:

- If the weight of people in a family is 30kg, 35kg, 40kg, 45kg . . .
- Rahul’s savings in Rs. for the first six months are 2000, 2500, 3000, 3500, 4000, 4500 . . .
- If the temperature (in Celsius) of an Indian State is measured 24°, 26°, 28°, 30°, 32°. . .

In all of the above examples, the common difference is constant.

You can **download the Formula pdf** Arithmetic Progression Class10 Formulas pdf Download.

**Truemaths** has all **NCERT video Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions.** In these video lectures, we have explained every detail about this chapter and the terms related to these chapters. We have also recorded NCERT Maths exercise questions for chapter 5. Visit **Truemaths** to get all the video solutions.

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In **Mathematics**, a series is called an * Arithmetic progression *or Arithmetic sequence if all the numbers in a series are placed in such a way that the difference between the consecutive terms is equal or constant. Here difference (also known as

Additionally outlining an Arithmetic Progression (AP) as a series which is increasing or decreasing by adding or subtracting a constant term to the previous term of the series.

**Arithmetic progression example:-**

The series of All-natural numbers are in AP as we are adding 1(constant) to write the next term.

(e.g.- 1,2,3,4,5………. )

The series of All even & odd numbers are in AP having the same common difference as 2.

Like in odd numbers here 1, 3, 5, 7, 9, 11, 13 . . . ; this Series is an arithmetic progression with common difference of two.

The easiest way to know, either the series is an AP or not, is finding the difference between the consecutive numbers. In Arithmetic Progression the difference between two consecutive terms is always constant or equal.

Or

Quantities in AP are always increasing or decreasing by a constant term also know as **common difference** denoted by “d”.

For example:

In an Even numbers series = 4 , 6 , 8,10 . . . .

In an Odd numbers series = 7 , 9 ,11 ,13 . . . .

So in the first series, we can find that the difference between consecutive term is constant i.e. +2.

Similarly, in the second series, the common difference is constant i.e. +2.

Ther are two main * Arithmetic progression formulas* which we will be discussing in length in this article. The first formula is to find any general term of an AP. Whereas with the help of second formula, we will be in a position to add n number of terms in an AP. In other words, the second formula is to find the sum of n terms of an AP.

First of all, let me explain how this formula is going to help you. Let us consider this very basic example. Assume that there is some boy named Rahul. Who wants his parents to give him some pocket money for his normal expenses. so he asked his parents to give him some pocket money. So his parents promised him that they will give him Rs. 100 per week. As Rahul thought this is very less so he again asked his parents to increase his pocket money and start giving his Rs. 1,000 per week instead.

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After some discussion, they all agreed that Rahul will not get Rs. 1,000 per week as his pocket money, instead what they will do they will start giving him, RS. 100 per week but they will increase his pocket money by Rs. 50 per week.

**Question 1**:- After how many weeks Rahul will get Rs. 1,000 as his pocket money?.

For this kind of questions, we need to form an AP and we need to find the nth term.

The general form of an Arithmetic Progression Series is given in the form of a, a + d, a + 2d, a + 3d…………a+(n-1)d. Thus the** formula to find the nth term** of an A.P series is given by.

a_{n}= a + ( n – 1 ) d

a_{n} = general term in the sequence.

a = the first in the sequence.

d = the common difference between any two consecutive terms.

some times, you will notice another formula. But both are same.

T_{n} = a + (n – 1) d

Where,

T_{n} = General term(a_{n}).

The remaining values are the same as in the first formula.

Series will be 100,150,200,250……….1000

a = 100

d = 50

a_{n} = 1000

n = ?

a_{n} = a+(n-1)d

1000 = 100 +(n-1)50

1000-100 = 50n-50

900+50 = 50n

950/50 = n

∴ n = 9

∴ After 9 weeks he will get Rs. 1000 as his pocket money.

In the last example let us consider another situation. Rahul wants to buy a cricket bat whose cost price is Rs. 3250. So obviously he will have to start saving his pocket money. So if we want to find after how many weeks will Rahul be able to save Rs. 3250 so that he can buy this bat. In the last situation, we already have an AP and in this situation, we have to find the sum of Rahul’s pocket money for some n weeks to find when he will be able to save that much money.

So in this situation, we want to add up the numbers in the sequence, To simplify this we use a formula that will help us to find the sum easily.

The **sum of the first n terms** of an AP is given by

S_{n} = n/2[2a + (n – 1) d ]

We can also write this as

S_{n}= n/2[a + a + (n – 1) d ]

S_{n} = n/2 (a + a_{n})

**let us solve this question.**

S_{n} = Rs. 3250

a = Rs.100

d = Rs. 50

n =?

3250 = n/2[ 2(100)+ (n-1)50]

⇒ 3250 = n/2[ 200+50n-50]

3250 = n/2 [150 +50n]

⇒3250 = n[75+25n]

n²+3n-130=0

n²+13n-10n-130 =0

n(n+13)-10(n+13) =0

(n+13)(n-10)=0

∴ n= -13,10

because number of weeks(n) can’t be negative, therefore number of weeks are 10.

- Finite and Infinite Arithmetic Progressions
- Ascending and descending Arithmetic Progression

- Finite Arithmetic Progression

If a series ends after sone terms, and there are a limited number of terms in the series then it is known as Finite Arithmetic Progression.

Examples: 122, 132 , 142 , 152 , 162

- Infinite Arithmetic Progression

If the series doesn’t end and there are an infinite number of terms within the series then it’s referred to as Infinite Arithmetic Progression.

Example: 4, 10 , 16 , 22 , 28 . . .

(Doted points means this series is continued till infinity)

- Ascending Arithmetic Progression

If the numbers in a series are in an increasing manner. It suggests that consecutive term is bigger than the previous term.

Eg. 13, 26, 39, 52, .. . . .

- Descending Arithmetic Progression

If the numbers in a series are in a decreasing manner. It suggests that consecutive term is smaller than the previous term.

Example: 40,30,20,10.. . . ..

- If A, B and C are in A.P. then B – A = C – B or 2B= A + C

For example: 10,15,20 . . . .

Here, A = 10; B = 15; C = 20 ;

So, we can apply above formula as 20-15 =15-10 = 5

2 X 15= 10 + 20 = 30

If a, b, c are in AP, b is the mid-term then b is the Arithmetic Mean for the series. AM can also found by the average of first and last term.

b = (a+b)/2

and b is called the arithmetic mean of a and c.

A collection of numbers arranged in a definite order according to some definite rule (rules) is named an Arithmetic Sequence.

The given series is a sequence:

a , a+d , a+2d , a+3d . . . . . . .

- Each number in the sequence is known as term. For example:

1000 , 2000 , 3000 , 4000 ,5000 . . . . . .so on.

In the above series, the difference between the numbers is equal i.e. 1000.

**Example:-**

Find the 10th term of the AP: 4,8,12,16 . . .

Solution

Given a = 4,

n = 10 (say we want to know 11th term of above AP)

d =8-4 = 4 (So we can say it’s a descending series)

an = a + (n – 1)d

Putting above values; a10 = 4+ (10-1) 4

= 4+(9)4

=4+36

= 40

The first and General Form of AP: | a, a + d, a + 2d, a + 3d, . . . |

And the formula Arithmetic Progression is: | |

The n^{th} term of AP | a_{n} = a + (n – 1) ×d |

and the sum formula of Arithmetic Progression: | |

Sum of an AP sequence | S_{n} = n/2[2a + (n − 1) × d] |

If Last term(l/a_{n} ) is given | S_{n}= n/2 (a+l) |

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