Arithmetic Progression: Properties, Formulas & Examples

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Arithmetic progression is defined as a pattern of sequence which has a typical relationship between two consecutive terms. It’s a series of numbers in which each term of the sequence diverges from the subsequent term by a continuing amount.

Arithmetic progression is also known as arithmetic sequence or AP.
It was reinvented by Carl Friedrich Gauss when he was in primary school.
The difference between two successive terms is called a common difference.
A sequence is defined as a finite or infinite list of numbers arranged in a certain format.
Finite AP and Infinite AP are two types of arithmetic progression.
It is used to calculate the nth term of AP and the sum of the nth term.
Roll numbers of students and days in a week are some real-life examples of arithmetic progression.

Table of Content

What is Arithmetic Progression?
Types of Arithmetic Progression
Properties of Arithmetic Progression
Notation in Arithmetic Progression
Arithmetic Progression Formula
Important Topics for JEE Main
Things to Remember
Sample Questions

Key Terms: Arithmetic Progression, Common Difference, First Term, Series, Sequence, Finite Arithmetic Progression, Infinite Arithmetic Progression, Sum of first n terms

What is Arithmetic Progression?

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Arithmetic progression (AP) is a sequence of series or numbers where the common difference between two consecutive terms is equal to constant. A progression is arranged in an exceedingly particular order such that the relation between consecutive terms of a series or sequence is usually constant.

Arithmetic progression is used to get a formula for the nth term.
In this, a new term is obtained by adding a continuing difference called a common difference.
It can be explained as a collection of integers within each term.
The terms are obtained by adding a constant to the preceding term aside from the primary term.
This means the second number is obtained by adding a constant to the first term.

Solved Example of Arithmetic Progression

Example: 1, 4, 7, 10, 13, 16, 19, 22, 25, ... is an arithmetic progression as the differences between every two consecutive terms are the same (as 3). i.e., 4 - 1 = 7 - 4 = 10 - 7 = 13 - 10 = 16 - 13 = 19 - 16 = 22 - 19 = 25 - 22 = ... = 3. We can also notice that every term (except the first term) of this AP is obtained by adding 3 to its previous term. In this arithmetic progression:

a = 1 (the first term)
d = 3 (the "common difference" between terms)

Arithmetic Progression Detailed Video Explanation:

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Types of Arithmetic Progression

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Arithmetic Progression is divided into two categories which are discussed in the following section:

Finite Arithmetic Progression

Finite Arithmetic Progression is a type of an arithmetic progression which consists of a finite number of terms.

Solved Example of Finite Arithmetic Progression Example 1: The example of finite arithmetic progression include 2,4,6,8,10 Example 2: The example of finite arithmetic progression include 3,6,9,12,15

Read More: Statistics

Infinite Arithmetic Progression

Infinite Arithmetic Progression is a type of an arithmetic progression which consists of an infinite number of terms.

Solved Example of Infinite Arithmetic Progression

Example 1: The examples of infinite arithmetic progression include 2,4,6,8,10,12,14,...

Example 2: The examples of infinite arithmetic progression include 3,6,9,12,15,18,21,...

Read More: Infinite Series

Properties of Arithmetic Progression

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Some important properties about arithmetic progression are as follows:

If a constant is added from every term of an AP, the resulting sequence is additionally an AP.
If each term of an AP is multiplied by a constant, the resulting sequence is additionally an AP.
If each term of an AP is divided by a non-zero constant, then the resulting sequence is additionally an arithmetic progression.
If a constant is subtracted from every term of an AP, the resulting sequence is additionally an AP.
Three numbers, a, b and c, are said to be in arithmetic progression if 2b = a + c.
The terms selected in regular intervals are said to be in arithmetic progression.

Read More: Arithmetic Geometric Sequence

Notation in Arithmetic Progression

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In arithmetic progression, some most important terms are as follows:

a= First Term
l= Last Term
d= Common Difference
n= Total Number Of Terms
an= n^th Term (where n=1,2,3,…,n)
Sn=Sum Of First n Terms

General Term of an Arithmetic Progression

Any sequence a₁ a₂ a₃…. and so on is termed in arithmetic progression if a_{n + 1}=a_n+d, n€N

Let’s assume the primary term of an arithmetic progression could be a, and the common difference is d i.e. a, a+d, a+2d, a+3d……

1^st term = a = a+(1-1) d

2^ndterm= a+d = a+(2-1) d

3^rd term = a+2d = a+(3-1) d.

N^thterm= an = a+(n-1) d

Read More: Sum of Squares

Arithmetic Progression Formula

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The important formula used in arithmetic progression are as follows:

Common Difference of an Arithmetic Progression

The formula for finding the n-th term of an arithmetic progression is as follows:

d = (a_{n + 1} – a_n) or d = (a_n – a_n-1)

where d is the common difference
a_n = n^th term
a_n+1 = n+1^th term
a_n-1=n-1^thterm

Solved Example of Common Difference of an Arithmetic Progression

Example 1: Find the value of d, if a = 10, n = 6, a_n = 95.

Solution: Given, a = 10, n = 6, a_n = 95

From the formula of general term, we have:

a_n = a + (n − 1) × d

95 = 10 + (6 − 1) × d

(6 − 1) × d = 95 – 10 = 85

d = 85/ 5

d = 17

Read More: Geometric Series Formula

n^th term of an Arithmetic Progression

The formula for calculating the nth term of arithmetic progression is as follows:

a_n = a + (n − 1)d

Where,
a = First term of AP
d = Common difference
n = number of terms
a_n = n^th term

Solved Example of n^th term of an Arithmetic Progression

Example 1: Find the nth term of AP: 1, 2, 3, 4, 5…., a_n, if the number of terms are 10.

Solution: Given, AP: 1, 2, 3, 4, 5…., a_n

n=10

By the formula we know, an = a+(n-1)d

First-term, a =1

Common difference, d=2-1 =1

Therefore, a_n = a₁₀ = 1+(10-1)1 = 1+9 = 10

Sum of n Terms of an Arithmetic Progression

The sum of nth terms of an arithmetic progression is the addition of the first n terms of the automated sequence. In simple words, the sum of n terms is divided by double of the sum of twice the primary term "a" and also the product of the difference between 2nd and, therefore, the first common difference.

The formula for finding the sum of n term of an arithmetic progression is as follows:

S = n/2 [2a + (n − 1) d]

Where,
a = First term of AP
d = Common difference
n = number of terms
In case when first and last terms are given then formula for the sum of n term of an arithmetic progression is as follows:

S = n/2 (first term of AP + last term of AP)

S = n/2[a+ a_n]

Where,
a = First term of AP
a_n = Last term of arithmetic progression
n = number of terms

Read More: Binomial Theorem

Solved Example of Sum of n Terms of an Arithmetic Progression

Example 1: Find the sum of the first 20 multiples of 5.

Solution: The first 20 multiples of 5 are 5, 10, 15, … 100.

Here, it is clear that the sequence formed is an arithmetic sequence where,
a = 5, d = 5, a_n = 100, n = 20.
S_n = n/2 [2a + (n − 1) d]
S_n = 20/2 [2 × 5 + (20 − 1)5]
S_n= 10 [10 + 95]
S_n = 1050

Example 2: Find the sum of the first 5 terms of the arithmetic progression whose first term is 2 and 5th term is 10.

Solution: We have a₁ = a = 2 and a₅ = 10 and n = 5.

Using the AP formula for the sum of n terms, we have
S_n = (n/2) (a + a_n)
S₅= (5/2) (2 + 10)
(5/2) × 12
30

The tabular representation of all formulas used in arithmetic progression are as follows:

Category	Formula
Common Difference	d = (a_n _{+ 1} – a_n) or d = (a_n – a_n-1)
General Form of Arithmetic Progression	a_n = a+(n-1) d
nth term of an Arithmetic Progression	a_n = a+(n-1) d
Sum of n terms of an Arithmetic Progression	n/2 [2a + (n − 1) d]
Sum of n terms of arithmetic progression when first and last term are given	n/2[a+ a_n]

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Important Topics for JEE Main

As per JEE Main 2024 Session 1, important topics included in the chapter arithmetic progression are as follows:

Types of AP
Properties of Arithmetic Progression
Arithmetic Progression Formula

Some memory based important questions asked in JEE Main 2024 Session 1 include:

Find p if: 3+ (3+p)/4+ (3 + 2p)/42 + ...∞ = 8
S₁ = 3,9,15, 25 terms and S2 = 3,8,13, ... 37 terms, then the number of common terms in S₁, S₂ is equal to?
Find the 20th term from the end of the progression: 20, 19(1/4), 18(1/2), 17(3/4), ..., -129(1/4)
If 2^nd, 8^th, 44^th term of A.P. is 1^st, 2^nd, and 3^rd terms respectively of G.P. and the first term of A.P. is 1, then the sum of the first 20 terms of A.P. is:
In an increasing arithmetic progression a₁, a₂,..., a_n if a₆ = 2 and the product of a₁, a₅ and a₄ is greatest, then the value of d is equal to?

Things to Remember

Arithmetic Progression is a sequence of numbers where the difference between two consecutive terms is equal to constant.
It is divided into two types, namely finite and infinite arithmetic progression.
The formula for the nth term of AP is a_n = a+(n-1) d.
The formula for common difference is given as d = (a_{n + 1} – a_n).
A sum of the first n terms of an arithmetic sequence is given as n/2 [2a + (n − 1) d].

Read More: Bayes Theorem Formula

Sample Questions

Ques: For what a value of K will K+9, 2k-1 and 2k+7 be consecutive terms of an A.P.? (2 marks)

Ans. (2k-1) - (k+9) = (2k+7) - (2k-1)

= k-10=8

= k=8

So, the value of K will be 8.

Ques: Find the “d” of an A.P. in which a21-a7=84? (2 marks)

Ans. a+(21-1)d - [a+(7-1)d] =84 [formula: tn=a+(n-1)d]

= (a+20d) - (a+6d) =84

= 14d=84

= d= 84/14 = 6

So, the common difference of an A.P. is 6.

Ques: Find the common difference of an A.P. whose first term is 4, the last term is 49 and the addition of all its terms is 265? (4 marks)

Ans. a=4, l=49, S=265

We know that, Sn=n/2(a+l)

Therefore, 265=n/2(4+49)

265=n/2(53)

n/2=265/53

n/2=5

n=10

So, the total number of terms is 10.

We know that, l=a+(n-1)d

Therefore, 49=4+(10-1)d

45=9d

d=45/9

d= 5

So, the common difference is 5.

Ques: The Addition of four consecutive numbers in an AP is 32 and the fraction of the multiple of the first and the last term to the multiple of two middle terms is 7:15. Find the numbers? (4 marks)

Ans. Let's suppose the four consecutive numbers are a-3d, a-d, a+d, a+3d

=The sum of four consecutive number is=32

=4a=32

=a=8

Furthermore,

(a-3d)(a+2d)/(a-d)(a+d)= a₂-9d₂/a₂-d₂=7/15

15a₂-135d₂=7a₂-7d₂

8a₂=128d₂

d₂=8/2

d₂=4

d= ±2

Now let's find out four consecutive numbers ,

a-3d=8-3(2)=2

a-d=8-2=6

a+d=8+2=10

a+3d=8+3(2)=14

So, the four consecutive numbers are 2,6,10,and14.

Ques: Find the addition of all factors of 7 lying between 500 and 900? (3 marks)

Ans. Our first term will be 504 and the last term will be 896.

We know that, l=a+(n-1)d

Therefore, 896=504+(n-1)7

392=(n-1)7

n-1=392/7=56

n=57

We know that, Sn=n/2(a+l)

Therefore, Sn=57/2(504+896)

=57/2×1400

=39900

Ques: If the ratio of the addition of the first n terms of two A.P’s is (7n+1):(4n+27), then find the fraction of their 9th terms? (2 marks)

Ans. Let's assume first term of (7n+1)= a₁ and common difference= d₁

Let's assume first term of (4n+27)= a₂ and common difference=d₂

So, our ratio is 24:19.

Ques: The sum of the first n terms of an A.P is given by S_n=3n²+2n. Determine the A.P and its 15th term? (4 marks)

Ans. We know that, Sn= 3n²+2n

First, let's find out S₁, S₂ and a₂,

S₁=a₁=3(1)²+2(1)

S₂=3(2)²+2(2)

=12+4

=16

a₂=S₂-S₁=16-5=11

So, d=a₂-a₁=11-5= 6

Now we will find out AP,

AP= a, a+d, a+2d,…….= 5, 11, 17,…..

And now we will see 15th term,

a₁₅=a+14d

=5+14(6)

=5+84=89

So, the common difference is 6 and its 15^th term is 89.

Ques: An operator of TV sets produced 600 sets in the 3rd year and 700 sets in the 7th year. Assuming that the production increases constantly by a fixed number every year, find the production in the 1st year, 10th year and the total production in the first 7 years? (4 marks)

Ans. Since the production increases constantly by a fixed number every year, the no. of TV sets operated in 1st,2nd,3rd,….., years will form an AP.

Let us denote the number of TV sets operator in the nth year by an.

Then, a₃= 600 and a₇= 700

or, a+2d=600

And a+6d=700

Solving these equations, we get d=25 and a=550.

Therefore production of TV sets in the 1st year is 550.

Now, a₁₀=a+9d

=550+9×25

=775

So, production of TV sets in the year is 775.

Also, S₇=7/2[2×550+(7-1)×25]

=7/2(1100+150)

=4375

Thus, the total production of TV sets in the first 7 years is 4375.

Ques: A man manages to pay off a loan of Rs 3600 by annual installments which form an arithmetic series. When 30 of the installments are paid, he dies leaving one-third of the loan unpaid. Find the value of the first installment? (5 marks)

Ans. A man manages to pay off a loan of Rs 3600 by 40 annual installments which form an A.P i.e., sum of all 40 installments= Rs 3600

S₄₀= 3600

We know that, S_n= n/2 [2a+(n-1)d]

40/x[2a+(40-1)d]=3600

2a+39d=3600/20=180….. (i)

He died by leaving one third of the loan unpaid. That means he paid the remaining money in 30 installments.

Therefore, the money he paid in 30 installments= 3600-3600/3=3600-1200

So, S₄₀=2400

S40=2400=30/2[2a+(30-1)d]=2400

Therefore, Sn=n/2[2a+(n-1)d]

2a+29d=2400/15=160…. (ii)

(i) – (ii) = 2a+39d=180

2a+29d=160/0+10d=20

d=20/10=2

Put d= 2 in (ii) 2a+29(2)=160

2a=102

a=102/2=51

Therefore, the value of his first installment = 51.

Ques: A man secured Rs 32 during the 1st year, Rs 36 in the 2nd year and in this way he raised his savings by Rs 4 every year. Find in what time his secured money will be Rs 200? (4 marks)

Ans. Saving in 1st year (a₁)=Rs 32

Saving in 2nd year (a₂)=Rs 36

Increase in salary every year (d)=Rs 4

Let in n years his saving will be Rs 200

= S_n=200

= n/2[2a+(n-1)d=200

= n/2(64+4n-4)=200

n/2(4n+60)=200

=2n²+30n=200

=n²+15-100=0 [when divided by 2]

=n²+20n-5n-100=0

=n(n+20)-5(n+20)=0

(n+20)(n-5)=0

If n+20=0 or n-5=0

n=-20 or n=5 [Rejected as n can’t be negative]

Therefore, in 5 years his savings will be Rs 200.

Ques: A man secured 16500 in 10 years. In each year after the first he secured Rs 100 more than he did in the previous year. How much did he secure in the 1st year? (4 marks)

Ans. Let “a” be the money he secured in 1st year

= First year he secured the money= Rs a

He saved Rs 100 more than he did in the previous year.

=Second year he secured the money=Rs (a+100)

= Third year he saved the money= Rs [a+2(100)]

So, the sequence is a, a+100, a+2(100), ……, This is an AP with common difference (d)=100

=Sum of money he secured in ten years S10=16,500 rupees

We know that, S_n=n/2[2a+(n-1)d]

S₁₀=10/2[2a+(10-1)100]

16500=5(2a+9×100)

2a +900=16500/5=3300

2a=2400

a=2400/2=1200

Therefore, he saved the money in the first year (a)= Rs 1200.

Ques. Which term of the AP 1, 4, 7, 10,... is 73? (3 marks)

Ans. The given progression is 1,4,7,10,...

Here the first term is a = 1, and the common difference is, d = 4 - 1= 7 - 4= ... = 3

Let us assume that the nth term is,

a_n = 78

Substitute all these values in the general term of an arithmetic progression:

a_n = a+(n - 1)d

73 = 1 +(n - 1)3

73 = 1 + 3n - 3

73 = 3n - 2

75 = 3n

15 = n

Ques. Find the AP if the first term is 12 and the common difference is 5? (3 marks)

Ans. As we know, a, a + d, a + 2d, a + 3d, a + 4d, …

Here, a = 12 and d = 5
12, (12 + 5), (12 + 2 × 5), (12 + 3 × 5), (12 + 4 × 5),
12, 17, (12 + 10), (12 + 15), (12 + 20), …
12, 17, 22, 27, 32, …and so on.
So the AP is 12, 17, 22, 27, 32…..

Ques: Find the common difference of an A.P. whose first term is 5, the last term is 51 and the addition of all its terms is 280? (3 marks)

Ans. a=5, l=51, S=280

We know that, S_n=n/2(a+l)

Therefore, 280=n/2(5+51)

280=n/2(56)

n/2=280/56

n/2=5

n=10

So, the total number of terms is 10.

We know that, l=a+(n-1)d

Therefore, 51=5+(10-1)d

46=9d

d=46/9

So, the common difference is 46/9.

Ques: Find the addition of all factors of 6 lying between 500 and 901? (3 marks)

Ans. Our first term will be 540 and the last term will be 900.

We know that, l=a+(n-1)d

Therefore, 900=540+(n-1)6

360=(n-1)6

n-1=360/6=60

n=61

We know that, Sn=n/2(a+l)

Therefore, Sn=61/2(540+900)

=61/2×1440

=43920

Ques. Mr. Kevin earns $400,000 per annum and his salary increases by $20,000 per annum. Then how much does he earn at the end of the first 2 years? (3 marks)

Ans. The amount earned by Mr. Kevin for the first year is, a = 4,00,000. The increment per annum is, d = 20,000. We have to calculate his earnings in the 2 years. So n = 2.

Substituting these values in the AP sum formula,

S_n = n/2 [2a + (n - 1) d]

S_n= 2/2(2(400000) + (2 - 1)(20000))

= 2/2 (800000 + 20000)

= 8,20,000

He earned $8,20,000 in 2 years.

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Arithmetic Progression: Properties, Formulas & Examples

What is Arithmetic Progression?

Solved Example of Arithmetic Progression

Arithmetic Progression Detailed Video Explanation:

Types of Arithmetic Progression

Finite Arithmetic Progression

Solved Example of Finite Arithmetic Progression

Infinite Arithmetic Progression

Solved Example of Infinite Arithmetic Progression

Properties of Arithmetic Progression

Notation in Arithmetic Progression

General Term of an Arithmetic Progression

Arithmetic Progression Formula

Common Difference of an Arithmetic Progression

Solved Example of Common Difference of an Arithmetic Progression

n^th term of an Arithmetic Progression

Solved Example of n^th term of an Arithmetic Progression

Sum of n Terms of an Arithmetic Progression

Solved Example of Sum of n Terms of an Arithmetic Progression

Important Topics for JEE Main

Things to Remember

Sample Questions

CBSE X Related Questions

1.
If \( 2 \sin A = 1 \), then the value of \( \tan A + \cot A \) is :

4.
Find the H.C.F. and L.C.M. of 408 and 312.

5.
Three coins are tossed together. The probability of getting exactly two tails is

6.
In the given figure, PQ is a tangent to a circle with centre \(O(-5, 3)\). If coordinates of P and Q are \((3, 1)\) and \((0, 6)\) respectively, then using distance formula, show that \(PQ \perp OQ\).

Similar Mathematics Concepts

Comments

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Arithmetic Progression: Properties, Formulas & Examples

What is Arithmetic Progression?

Solved Example of Arithmetic Progression

Arithmetic Progression Detailed Video Explanation:

Types of Arithmetic Progression

Finite Arithmetic Progression

Solved Example of Finite Arithmetic Progression

Infinite Arithmetic Progression

Solved Example of Infinite Arithmetic Progression

Properties of Arithmetic Progression

Notation in Arithmetic Progression

General Term of an Arithmetic Progression

Arithmetic Progression Formula

Common Difference of an Arithmetic Progression

Solved Example of Common Difference of an Arithmetic Progression

nth term of an Arithmetic Progression

Solved Example of nth term of an Arithmetic Progression

Sum of n Terms of an Arithmetic Progression

Solved Example of Sum of n Terms of an Arithmetic Progression

Important Topics for JEE Main

Things to Remember

Sample Questions

CBSE X Related Questions

1.If \( 2 \sin A = 1 \), then the value of \( \tan A + \cot A \) is :

4.Find the H.C.F. and L.C.M. of 408 and 312.

5.Three coins are tossed together. The probability of getting exactly two tails is

6.In the given figure, PQ is a tangent to a circle with centre \(O(-5, 3)\). If coordinates of P and Q are \((3, 1)\) and \((0, 6)\) respectively, then using distance formula, show that \(PQ \perp OQ\).

Similar Mathematics Concepts

Comments

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n^th term of an Arithmetic Progression

Solved Example of n^th term of an Arithmetic Progression

1.
If \( 2 \sin A = 1 \), then the value of \( \tan A + \cot A \) is :

4.
Find the H.C.F. and L.C.M. of 408 and 312.

5.
Three coins are tossed together. The probability of getting exactly two tails is

6.
In the given figure, PQ is a tangent to a circle with centre \(O(-5, 3)\). If coordinates of P and Q are \((3, 1)\) and \((0, 6)\) respectively, then using distance formula, show that \(PQ \perp OQ\).