**Arithmetic Sequence**

A sequence *a*_{1}, *a*_{2}, *a*_{3}, …, *a _{n}*, …

is called an

**arithmetic sequence**, or

**arithmetic progression**, if there exists a constant

*d*, called the

**common difference**, such that

*a*–

_{n}*a*

_{(n-1)}=

*d*

That is,

*a*–

_{n}*a*

_{(n-1)}=

*d*for every

*n*1.

FOCUS: Relate linear functions and arithmetic sequences, then solve problems related to arithmetic sequences.

In an arithmetic sequence, the difference between consecutive terms is constant. This constant value is called the **common difference**.

This is an arithmetic sequence:

The first term of this sequence is:

*t*

_{1}=4 The second term is:

*t*

_{2}=7

Let

*d*represent the common difference. For the sequence above:

The dots indicate that the sequence continues forever; it is an

**infinite arithmetic sequence**.

To graph this arithmetic sequence, plot the term value,

*t*, against the term number,

_{n}*n*.

**Graph at an Arithmetic Sequence**

The graph represents a linear function because the points lie on a straight line. A line through the points on the graph has slope 3, which is the common difference of the sequence.

In an arithmetic sequence, the common difference can be any real number.

Here are some other examples of arithmetic sequences.

▪ This is an *increasing* arithmetic sequence because *d* is positive and the terms are decreasing:

*d*=¼

▪ This is a

*decreasing*arithmetic sequence because

*d*is negative and the terms are decreasing:

*d*=-6

An arithmetic progression is a list of numbers where the difference between successive numbers is constant. The terms in an arithmetic progression are usually denoted as *u*_{1}, *u*_{2}, *u*_{3} etc. where ul is the initial term in the progression, *u*_{2} is the second term, and so on; *u _{n}* is the

*n*th term. An example of an arithmetic progression is

Since the difference between successive terms is constant, we have

*u*

_{3}–

*u*

_{2}=

*u*

_{2}–

*u*

_{1}

and in general

*u*

_{(n+1)}–

*u*=

_{n}*u*

_{2}–

*u*

_{1}

We will denote the difference

*u*

_{2}–

*u*

_{1}as

*d*, which is a common notation.

Now, to know about an AP, what is the minimum information that you need? Is it enough to know the first term? Or, is it enough to know only the common difference? You will find that you will need to know both — the first term *a* and the common difference *d*.

For instance if the first term *a* is 6 and the common difference *d* is 3, then the AP is

and if a is 6 and

*d*is -3, then the AP is

Similarly, when

*a*=-7,

*d*=-2, the AP is -7, -9, -11, -13, …

*a*=1.0,

*d*=0.1, the AP is 1.0, 1.1, 1.2, 1.3, …

*a*=0,

*d*=1½, the AP is 0, 1½, 3, 4½, 6, …

*a*=2,

*d*=0, the AP is 2, 2, 2, 2, …

So, if you know what

*a*and

*d*are, you can list the AP. What about the other way round? That is, if you are given a list of numbers can you say that it is an AP and then find

*a*and

*d*? Since a is the first term, it can easily be written. We know that in an AP, every succeeding term is obtained by adding

*d*to the preceding term. So,

*d*found by subtracting any term from its succeeding term, i.e., the term which immediately follows it should be same for an AP.

For example, for the list of numbers: 6, 9, 12, 15, …,

We have

*a*

_{2}–

*a*

_{1}=9-6=3,

*a*

_{3}–

*a*

_{2}=12-9=3,

*a*

_{4}–

*a*

_{3}=15-12=3.

Here the difference of any two consecutive terms in each case is 3. So, the given list is an AP whose first term

*a*is 6 and common difference

*d*is 3.

For the list of numbers: 6, 3, 0, -3, …,

*a*

_{2}–

*a*

_{1}=3-6=-3,

*a*

_{3}–

*a*

_{2}=0-3=-3,

*a*

_{4}–

*a*

_{3}=-3-0=-3.

Similarly this is also an AP whose first term is 6 and the common difference is -3.

In general, for an AP

*a*

_{1},

*a*

_{2}, …,

*a*, we have

_{n}*d*=

*a*

_{(k+1)}–

*a*

_{k}where

*a*

_{(k+1)}and

*a*are the (

_{k}*k*+1)th and the

*k*th terms respectively.

To obtain *d* in a given AP, we need not find all of *a*_{2}–*a*_{1}, *a*_{3}–*a*_{2}, *a*_{4}–*a*_{3}, …. It is enough to find only one of them.

Consider the list of numbers 1, 1, 2, 3, 5, … . By looking at it, you can tell that the difference between any two consecutive terms is not the same. So, this is not an AP.

Note that to find *d* in the AP: 6, 3, 0, -3, …, we have subtracted 6 from 3 and not 3 from 6, i.e., we should subtract the *k*th term from the (*k*+1) th term even if the (*k*+1) th term is smaller.

An

arithmetic sequenceis a sequence in which each term differs from the previous one by the same fixed number.

It can also be referred to as anarithmetic progression.

For example:

● the tower of bricks in the previous section forms an arithmetic sequence where the difference between terms is 1

● 2, 5, 8, 11, 14, … is arithmetic as 5-2=8-5=11-8=14-11=….

● 31, 27, 23, 19, … is arithmetic as 27-31=23-27=19-23=….

**ALGEBRAIC DEFINITION**

{

u} is_{n}arithmetic⇔u_{(n+1)}–u=_{n}dfor all positive integersnwheredis a constant called thecommon difference.

The symbol ⇔ means ‘if and only if’. It implies both:

● if {*u _{n}*} is arithmetic then

*u*

_{(n+1)}–

*u*is a constant

_{n}● if

*u*

_{(n+1)}–

*u*is a constant then {

_{n}*u*} is arithmetic.

_{n}Let us make the concept more clear through some examples.

Example 1: Find the first term *a*nd the common difference of each of the following arithmetic progressions.

(i) 7, 11, 15, 19, 23, …

(ii) ⅙, ⅓, ½, ⅔, ⅚, 1, …

(iii) *a*+2*b*, *a*+*b*, *a*, a-*b*, a-2*b*, …

Solution:

First term | Common difference | |
---|---|---|

(i) | 7 | 4 |

(ii) | ⅙ | ⅓ |

(iii) | a+2b |
–b |

Example 2: Given that 3, 7 and 11 are the first three terms in an arithmetic progression, what is *d*?

Then

*d*=4. That is, the common difference between the terms is 4.

Example 3: Given that 2*x*, 5 and 6-*x* are the first three terms in an arithmetic progression, what is *d*?

*x*=(6-

*x*)-5

5-2

*x*=1-

*x*

4=2

*x*–

*x*

*x*=4

Since

*x*=4, the terms are 8, 5, 2 and the difference is -3. The next term in the arithmetic progression will be -1.

Ex4. For the following APs, write the first term *a*nd the common difference:

(i) 3, 1, -1, -3, … (ii) -5, -1, 3, 7 …

(iii) ⅓, 5/3, 9/3, 13/3 … (iv) 0.6, 1.7, 2.8, 3.9 …

Solution:

(i) First term *a*=3

Common difference *d*=*a*_{2}–*a*_{1}=1-3=-2

(ii) First term *a*=-5

Common difference *d*=*a*_{2}–*a*_{1}=-1-(-5)=4

(iii) First term *a*=⅓

Common difference *d*=*a*_{2}–*a*_{1}=5/3-⅓=4/3

(iv) First term *a*=0.6

Common difference *d*=*a*_{2}–*a*_{1}=1.7-0.6=1.1

Question 10:

The 17th term of an AP exceeds its 10th term by 7. Find the common difference.

Answer 10:

Let the first term=*a* and common difference =*d*

According to question, *a*_{17}=*a*_{10}+7

*a*+16

*d*=

*a*+9d+7

7

*d*=7

*d*=1

Hence, the common difference is 1.

Ex5. For the AP: 3/2, ½, -½. -3/2, …, write the first term *a* and the common difference *d*.

Solution: Here, *a*=3/2, *d*=½-3/2=-1.

Remember that we can find *d* using any two consecutive terms, once we know that the numbers are in AP.

Ex6. For the following arithmetic progressions write the first term *a* and the common difference d:

(i) -5, -1, 3, 7, …

(ii) ⅕, ⅗, 5/5, 7/5, …

(iii) 0.3, 0.55, 0.80, 1.05, …

(iv) -1.1, -3.1, -5.1, -7.1, …

Solution:

We know that if a is the first term *a*nd *d* is the common difference, the arithmetic progression is *a*, *a*+*d*, *a*+2*d*, *a*+3*d*, … .

(i) -5, -1, 3, 7, …

Given arithmetic progression is -5, -1, 3, 7, …

This is in the form of *a*, *a*+*d*, *a*+2*d*, *a*+3*d*, … by comparing these two *a*=-5, *a*+*d*=1, *a*+2*d*=3, *a*+3*d*=7, ….

First term (*a*)=-5

By subtracting second and first term, we get (*a*+*d*)-(*a*)=*d*.

*d*

4=

*d*

Common difference (

*d*)=4.

(ii) ⅕, ⅗, 5/5, 7/5, …

*a*, *a*+*d*, *a*+2*d*, *a*+3*d*, …. By comparing this two, we get

*a*=⅕,

*a*+

*d*=⅗,

*a*+2

*d*=5/5,

*a*+3

*d*=7/5, ….

First term

*a*=⅕.

By subtracting first term from second term, we get

*d*=(

*a*+

*d*)-(

*a*)

*d*=⅗-⅕=⅖

common difference

*d*=⅖.

iii) 0.3, 0.55, 0.80, 1.05, …

Given arithmetic progression, 0.3, 0.55, 0.80, 1.05, …

General arithmetic progression *a*, *a*+*d*, *a*+2*d*, *a*+3*d*, ….

By comparing,

*a*=0.3,

*a*+

*d*=0.55,

*a*+2

*d*=0.80,

*a*+3

*d*=1.05.

First term (

*a*)=0.3.

By subtracting first term from second term. We get

*d*=(

*a*+

*d*)-(

*a*)

*d*=0.55-0.3

*d*=0.25

Common difference (

*d*)=0.25.

(iv) -1.1, -3.1, -5.1, -7.1, …

General arithmetic progression is *a*, *a*+*d*, *a*+2*d*, *a*+3*d*, …

By comparing this two, we get *a*=-1.1, *a*+*d* =-3.1, *a*+2d =-5.1, *a*+3d =-7.1.

First term (*a*)=-1.1

Common difference (*d*)=(*a*+*d*)–(*a*)

Common difference (

*d*)=-2

Example 7: Arithmetic sequences

Show that the sequence is arithmetic; find the common difference and the twentieth term.

(a) *a _{n}*=2

*n*-1 (b) 50, 45, 40, …, 55-5

*n*, …

Solution:

(a) The first few terms of {

*a*} are 1, 3, 5, 7, . . . , from which it is apparent that each term is 2 more than the preceding term; this is an arithmetic sequence with first term

_{n}*a*nd common difference

*a*=l and

*d*=2. Check to see that

*a*

_{(n+1)}–

*a*=2. To find

_{n}*a*

_{20}, use either the defining formula for the sequence or Equation (1) for the

*n*th term:

*a*

_{20}=2⋅20-1=39 or

*a*

_{20}=

*a*+19

*d*=1+19⋅2=39.

(b) If

*b*=55-5

_{n}*n*, then

*b*

_{(n+1)}–

*b*=[55-5(

_{n}*n*+1)]-[55-5

*n*(

*n*+1)]=-5. This is an arithmetic sequence with

*a*=50,

*d*=-5, and so

*b*

_{20}=55-5⋅20=-45.

Ex8. In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?

(i) The taxi fare after each km when the fare is $15 for the first km and $8 for each additional km.

(iii) The cost of digging a well after every metre of digging, when it costs $150 for the first metre and rises by $50 for each subsequent metre.

Solution:

(i) Fare for 1 km *a*_{1}=$15

Fare for 2 km *a*_{2}=$(15+8)=$23

Fare for 3 km *a*_{3}=$(23+8)=$31

Fare for 4 km *a*_{4}=$(31+8)=$39

*a*

_{2}–

*a*

_{1}=23-15=8

*a*

_{3}–

*a*

_{2}=31-23=8

*a*

_{4}–

*a*

_{3}=39-31=8

The difference between the successive terms are same. Hence, it is an A.P.

(iii) Cost for digging 1 m deep *a*_{1}=$150

Cost for digging 2 In deep *a*_{2}=$(150+50)=$200

Cost for digging 3 in deep *a*_{3}=$(200+50)=$250

Cost for digging 4 In deep *a*_{4}=$(250+50)=$300

*a*

_{2}–

*a*

_{1}=200-150=50

*a*

_{3}–

*a*

_{2}=250-200=50

*a*

_{4}–

*a*

_{3}=300-250=50

The difference between the successive terms are same. Hence, it is an A.P.

Ex9. Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?

Solution:

Let the first term of the first AP =A and the common difference =*d*.

Let the first term of the second AP =*a* and the common difference =*d*.

Difference between their 100th term =*a*_{100}–*a*_{100}=100

*A*+99

*d*)-(

*a*+99

*d*)=100

*A*–

*a*=100

Difference between their 1000th term=

*a*

_{1000}–

*a*

_{1000}=100

*A*+999

*d*)-(

*a*+999

*d*)=100

*A*–

*a*=100

Hence, the difference between their 1000th terms is 100.

Ex10. Show that the sequence defined by *a _{n}*=5

*n*-7 is an A.P., find its common difference.

Solution:

Given sequence is

*a*=5

_{n}*n*-7

*n*th term of given sequence (

*a*)=5

_{n}*n*-7.

(

*n*+1)th term of given sequence

*a*

_{(n+1)}.

*a*

_{(n+1)}=5(

*n*+1)-7=5

*n*+5-7

*a*

_{(n+1)}=5

*n*-2

Hence, the common ration is given by (

*a*

_{(n+1)}–

*a*).

_{n}*a*

_{(n+1)}–

*a*=(5

_{n}*n*-2)-(5

*n*-7)

=5

∴

*d*=5

Ex11. Show that the sequence defined by *a _{n}*=3

*n*

^{2}-5 is not an A.P.

Solution:

Given sequence is,

*a*=3

_{n}*n*

^{2}-5.

*n*th term of given sequence (

*a*) =3⋅

_{n}*n*

^{2}-5.

(

*n*+1)th term of given sequence (

*a*

_{(n+1)})=3⋅(

*n*+1)

^{2}-5

*n*

^{2}+2n⋅1+1

^{2})-5

=3

*n*

^{2}+6n-2

The common difference (

*d*)=

*a*

_{(n+1)}–

*a*.

_{n}*d*=(3

*n*

^{2}+6n-2)-(3

*n*

^{2}-5)

=3

*n*

^{2}+6n-2-3

*n*

^{2}+5

=6

*n*+3

Common difference (

*d*) depends on ‘

*n*‘ value, ∴ given sequence is not in A.P.