A number sequence is an ordered list of numbers defined by a rule.
The numbers in the sequence are said to be its members or its terms.
A sequence which continues forever is called an infinite sequence.
A sequence which terminates is called a finite sequence.

For example, 3, 7, 11, 15, … form an infinite number sequence.
The first term is 3, the second term is 7, the third term is 11, and so on.
We can describe this pattern in words:

“The sequence starts at 3 and each term is 4 more than the previous term.”

Thus, the fifth term is 19 and the sixth term is 23.
The sequence 3, 7, 11, 15, 19, 23 which terminates with the sixth term, is a finite number sequence.

Example 1. Describe the sequence: 14, 17, 20, 23, and write down the next two terms.
solution:
The sequence starts at 14, and each term is 3 more than the previous term.
The next two terms are 26 and 29.

Arithmetic sequences
Arithmetic sequence is a sequence where the common difference (d) between consecutive terms is constant.

a₂–a₁=a₃–a₂=a_n–a_(n-1)=d (common difference)

Example 2. Given the sequence: 5, 9, 13, 17, … .
a) Determine the common difference
b) Determine the next two terms
Solution:

d=9-5=13-9=4
a₅=17+4=21 and a₆=21+4=25

Worked Example: Find the Next Terms
Find the next four terms of the arithmetic sequence 55, 49, 43, … .

Find the common difference d by subtracting two consecutive terms.
49-55=-6 and 43-49=-6 So, d=-6.
Now add -6 to the third term of the sequence, and then continue adding -6 until the next four terms are found.

The next four terms of the sequence are 37, 31, 25, and 19.

It is possible to develop a formula for each term of an arithmetic sequence in terms of the first term a₁ and the common difference d. Consider Example 1.

The following formula generalizes this pattern for any arithmetic sequence.

Key Concept: nth Term of an Arithmetic Sequence
The nth term a_n of an arithmetic sequence with first term a₁ and common difference d is given by

a_n=a₁+(n-1)d.

Example 3. Which of the following list of numbers form an AP? If they form an AP, write the next two terms:
(i) 4, 10, 16, 22, …
(ii) 1, -1, -3, -5, …
(iii) -2, 2, -2, 2, -2, …
(iv) 1, 1, 1, 2, 2, 2, 3, 3, 3, …
Solution:
(i) We have

a₂–a₁=10-4=6
a₃–a₂=16-10=6
a₄–a₃=22-16=6
i.e., a_(k+1)–a_k is the same every time.
So, the given list of numbers forms an AP with the common difference d=6.
The next two terms are: 22+6=28 and 28+6=34.

(ii) a₂–a₁=-1-1=-2

a₃–a₂=-3-(-1)=-3+1=-2
a₄–a₃=-5-(-3)=-5+3=-2
i.e., a_(k+1)–a_k is the same every time.
So, the given list of numbers forms an AP with the common difference d=-2.
The next two terms are:
-5+(-2)=-7 and -7+(-2)=-9

(iii) a₂–a₁=2-(-2)=2+2=4

a₃–a₂=-2-2=-4
As a₂–a₁≠a₃–a₂, the given list of numbers does not form an AP.

(iv) a₂–a₁=1-1=0

a₃–a₂=1-1=0
a₄–a₃=2-1=1
Here, a₂–a₁=a₃–a₂≠a₄–a₃.
So, the given list of numbers does not form an AP.

Question 1. Which of the following are APs? If they form an AP, find the common difference d and write three more terms.

Solution:
(i) 2, 4, 8, 16, …

a₂–a₁=4-2=2
a₃–a₂=8-4=4
a₄–a₃=16-8=8
The difference between the successive terms are not same. Hence, it is not an A.P.

(ii) 2, 2½, 3, 3½, …

a₂–a₁=2½-2=½
a₃–a₂=3-2½=½
a₄–a₃=3½-3=½
The difference between the successive terms are same. Hence, it is an A.P.
Common difference =½, next three terms of this AP is as follows:
Fifth term a₅=a₄+d=7½+½=4
Sixth term a₆=a₅+d=4+½=4½
Seventh term a₇=a₆+d=4½+½=5

(iii) -1.2, -3.2, -5.2, -7.2, …

a₂–a₁=-3.2-(-1.2)=-2.0
a₃–a₂=-5.2-(-3.2)=-2.0
a₄–a₃=-7.2-(-5.2)=-2.0
The difference between the successive terms are same. Hence, it is an A.P.
Common difference =-2.0, next three terms of this AP is as follows:
Fifth term a₅=a₄+d=-7.2-2.0=-9.2
Sixth term a₆=a₅+d=-9.2-2.0=-11.2
Seventh term a₇=a₆+d=-11.2-2.0=-13.2

(iv) -10, -6, -2, 2, …

a₂–a₁=-6-(-10)–4
a₃–a₂=-2-(-6)=4
a₄–a₃=2-(-2)=4
The difference between the successive terms are same. Hence, it is an A.P. Common difference =4, next three terms of this AP is as follows:
Fifth term a₅=a₄+d=2+4=6
Sixth term a₆=a₅+d=6+4=10
Seventh term a₇=a₆+d=10+4=14

(v) 3, 3+√2, 3+2√2, 3+3√2, …

a₂–a₁=3+√2-3=√2
a₃–a₂=3+2√2-(3+√2)=√2
a₄–a₃=3+3√2-(3+2√2)=√2
The difference between the successive terms are same. Hence, it is an A.P. Common difference =4, next three terms of this AP is as follows:
Fifth term a₅=a₄+d=3+3√2+√2=3+4√2
Sixth term a₆=a₅+d=3+4√2+√2=3+5√2
Seventh term a₇=a₆+d=3+5√2+√2=3+6√2

(vi) 0.2, 0.22, 0.222, 0.2222, …
a₂–a₁=0.22-0.2=0.02
a₃–a₂=0.222-0.22=0.002
a₄–a₃=0.2222-0.222=0.0002
The difference between the successive terms are not same. Hence, it is not an A.P.

(vii) 0, -4, -8, -12, …

a₂–a₁=-4-0=-4
a₃–a₂=-8-(-4)=-4
a₄–a₃=-12-(-8)=-4
The difference between the successive terms are same. Hence, it is an A.P. Common difference=-4, next three terms of this AP is as follows:
Fifth term a₅=a₄+d=-12-4=-16
Sixth term a₆=a₅+d=-16-4=-20
Seventh term a₇=a₆+d=-20-4=-24

(viii) -½, -½, -½, -½, …

a₂–a₁=-½-(-½)=0
a₃–a₂=-½-(-½)=0
a₄–a₃=-½-(-½)=0
The difference between the successive terms are same. Hence, it is an A.P.
Common difference =0, next three terms of this AP is as follows:
Fifth term a₅=a₄+d=-½+0=-½
Sixth term a₆=a₅+d=-½+0=-½
Seventh term a₇=a₆+d-½+0=-½

(ix) 1, 3, 9, 27, …

a₂–a₁=3-1=2
a₃–a₂=9-3=6
a₄–a₃=27-9=18
The difference between the successive terms are not same. Hence, it is not an A.P.

(x) a, 2a, 3a, 4a, …

a₂–a₁=2a–a=a
a₃–a₂=3a-2a=a
a₄–a₃=4a-3a=a
The difference between the successive terms are same. Hence, it is an A.P. Common difference=a, next three terms of this AP is as follows:
Fifth term a₅=a₄+d=4a+a=5a
Sixth term a₆=a₅+d=5a+a=6a
Seventh term a₇=a₆+d=6a+a=7a

(xi) a, a², a³, a⁴, …

a₂–a₁=a²-a
a₃–a₂=a³-a²
a₄–a₃=a⁴-a³
The difference between the successive terms are not same. Hence, it is not an A.P.

(xii) √2, √8, √18 , √32, …

a₂–a₁=√8-√2=2√2-√2=√2
a₃–a₂=√18-√8=3√2-2√2=√2
a₄–a₃=√32-√18=4√2-3√2=√2
The difference between the successive terms are same. Hence, it is an A.P. Common difference =√2 next three terms of this AP is as follows:
Fifth term a₅=a₄+d=4√2+√2=5√2=√50
Sixth term a₆=a₅+d=5√2+√2=6√2=√72
Seventh term a₇=a₆+d=6√2+√2=7√2=√98

(xiii) √3, √6, √9 , √12, …

a₂–a₁=√6-√3=√3 (√2-1)
a₃–a₂=√18-√8=√9-√6=√3 (√3-√2)
a₄–a₃=√32-√18=√12 -√9=√3 (√4-√3)
The difference between the successive terms are not same. Hence, it is not an A.P.

(xiv) 1², 3², 5², 7², …

a₂–a₁=3²-1²=9-1=8
a₃–a₂=5²-3²=25-9=16
a₄–a₃=7²-5²=49-25=24
The difference between the successive terms are not same. Hence, it is not an A.P.

(xv) 1², 5², 7², 73, …

a₂–a₁=5²-1²=25-1=24
a₃–a₂=7²-5²=49-25=24
a₄–a₃=73-7²=73-49=24
The difference between the successive terms are same. Hence, it is an A.P.
Common difference =24, next three terms of this AP is as follows:
Fifth term a₅=a₄+d=73+24=97
Sixth term a₆=a₅+d=97+24=121
Seventh term a₇=a₆+d=121+24=145