# How to calculate the number of Terms of an Arithmetic Sequence?

Arithmetic sequences
We will limit our attention for the moment to one particular type of sequence, known as an arithmetic sequence (or arithmetic progression (AP)). This is a sequence of the form

a, a+d, a+2d, a+3d, …

where each term is obtained from the preceding one by adding a constant, called the common difference and often represented by the symbol d. Note that d can be positive, negative or zero.

Thus, the sequence of even numbers

2, 4, 6, 8, 10, …

is an arithmetic sequence in which the common difference is d=2.
It is easy to see that the formula for the nth term of an arithmetic sequence is
an=a+(n-1)d.

Example 1. Find the number of terms in each of the following APs:
(i) 7, 13, 19, …, 205 (ii) 18, 15½, 13, …, -47
Solution:
(i) Here, a=7 and d=13-7=6.
Let the total number of terms in the AP is n.
Therefore, an=205

a+(n-1)d=205
7+(n-1)(6)=205
(n-1)(6)=198
n-1=33
n=34

Hence, there are 34 terms in the given AP.

(ii) Here, a=18 and d=15½-18=-2½.
Let the total number of terms in the AP is n.
Therefore, an=-47
a+(n-1)d=-47

18+(n-1)(-2½)=-47
(n-1) (-2½)=-65
n-1=26
n=27

Hence, there are 27 terms in the given AP.

Ex2. The first term of an arithmetic sequence is 5, the common difference is 3 and the last term is 80, find the number of terms.
Solution:
Given
First term (a)=5
Common difference (d)=3
Last term ()=80
To calculate no of terms in given arithmetic sequence an=a+(n-1)d
Let an=80,

80=5+(n-1)⋅3
75÷3=n-1
n=26

∴ There are 26 terms.

Question 1. In a flower bed, there are 23 rose plants in the first row, 21 in the second, 19 in the third, and so on. There are 5 rose plants in the last row. How many rows are there in the flower bed?
Solution:
The number of rose plants in the 1st, 2nd, 3rd, …, rows are:

23, 21, 19, …, 5

It forms an AP (Why?). Let the number of rows in the flower bed be n.
Then a=23, d=21-23=-2, an=5
As, an=a+(n-1)d
We have, 5=23+(n-1)(-2)
-18=(n-1)(-2)
n=10

So, there are 10 rows in the flower bed.

Question 2. How many two-digit numbers are divisible by 3?
Solution:
The list of two-digit numbers divisible by 3 is:

12, 15, 18, …, 99

Is this an AP? Yes it is. Here, a=12, d=3, an=99.
an=a+(n-1)d
99=12+(n-1)×3
87=(n-1)×3
n-1=87÷3=29
n=29+1=30

So, there are 30 two-digit numbers divisible by 3.

Question 3. How many three-digit numbers are divisible by 7?
Solution:
Three digit numbers divisible by 7: 105, 112, 119, …, 994.
Let the total number of these numbers be n.
Here, a=105 and d=112-105=7 to find n, where an=994. Given that: an=a+(n-1)d=994

⇒105+(n-1)(7)=994
7(n-1)=889
n-1=889/7=127
n=128

Hence, there are 128 three digits numbers which are divisible by 7.

Question 4. How many multiples of 4 lie between 10 and 250?
Solution:
Multiples of 4 lie between 10 and 250: 12, 16, 20, …, 248.
Let the total number of multiples of 4 lie between 10 and 250 be n.
Here, a=12 and d=16-12=4 to find n, where an=248.
Given that: an=a+(n-1)d=248

⇒12+(n-1)(4)=248
⇒4(n-1)=236
n-1=236/4=59
n=60

Hence, the total number of multiples of4 lie between 10 and 250 is 60.

Q5. How many terms are there in the arithmetic sequence?
(i) 7, 10, 13, …, 43.
(ii) -1, -⅚, -⅔, -½, …, 10/3
(iii) 7, 13, l9, …, 205
(iv) 18, 15½, 13, …, -47
Solution:
(i) 7, 10, 13, …, 43.
From given arithmetic sequence

a=7, d=10-7=3, an=a+(n-1)d.

Let, an=43 (last term)
7+(n-1)⋅3=43
n-1=36/3=12
n=13

∴ 13 terms are there in given arithmetic sequence

(ii) -1, -⅚, -⅔, -½, …, 10/3
From given arithmetic sequence

a=-1, d=-⅚+1=⅙, an=a+(n-1)d

Let, an=10/3 (last term)
-1+(n-1)⅙=10/3
n-1=(10/3+1)⋅6
n-1=20+6
n=27

∴ 27 terms are there in given arithmetic sequence

(iii) 7, 13, 19, …, 205
From the given arithmetic sequence
a=7, d=13-7=6, an=a+(n-1)d
Let, an=205 (last term)

7+(n-1)⋅6=205
(n-1)⋅6=198
n-1=33
n=34

∴ 34 terms are there in given arithmetic sequence

(iv) 18, 15½, 13, …, -47
From the given arithmetic sequence.,

a=18, d=15½-18=15.5-18=-2.5
an=a+(n-1)d

Let an=-47 (last term)
18+(n-1)⋅(-2.5)=-47
18+47=2.5(n-1)
n-1=65÷2.5
n-1=26
n=27
∴ 27 terms are there in given arithmetic sequence

Q6. How many terms are there in the arithmetic sequence 6, 10, 14, 18, …, 174?
Solution:
In the given arithmetic sequence, a=6 and d=(10-6)=4.
Suppose that there are n terms in the given arithmetic sequence.
Then, an=174

a+(n-1)d=174
6+(n-1)⋅4=174
2+4n=174
4n=172
n=43.

Hence, there are 43 terms in the given arithmetic sequence.

Q7. How many terms are there in the arithmetic sequence 41, 38, 35, …, 8?
Solution:
In the given arithmetic sequence, a=41 and d=(38-41)=-3.
Suppose that there are n terms in the given arithmetic sequence. Then an=8

a+(n-1)d=8
41+(n-1)⋅(-3)=8
44-3n=8
3n=36
n=12.

Hence, there are 12 terms in the given arithmetic sequence.

Q8. How many terms are there in the arithmetic sequence 18, 15½, 13, …, -47?
Solution:
The given arithmetic sequence is 18, 15½, 13, …, -47.
First term, a=18, common difference, d=15½-18=-2½=-5/2
Suppose there are n terms in the given arithmetic sequence. Then,

an=-47
18+(n-1)⋅(-5/2)=-47
-5/2 (n-1)=-47-18
n-1=-65⋅(-⅖)
n-1=26
n=27.

Hence, there are 27 terms in the given arithmetic sequence.

Q9. In a flower bed, there are 43 rose plants in the first row, 41 in second, 39 in the third, and so on. There are 11 rose plants in the last row. How many rows are there in the flower bed?
Solution:
The numbers of rose plants in consecutive rows are 43, 41, 39, …, 11.
Difference of rose plants between two consecutive rows =(41-43)=(39-41)=-2 [Constant].
So, the given progression is an arithmetic sequence
Here, first term =43, common difference =-2, last term =11.
Let this arithmetic sequence contains n terms, then we have

an=a+(n-1)d
11=43+(n-1)(-2)
11=45-2n
34=2n
n=17

Hence, the 17th term is 11 or there are 17 rows in the flower bed.
Let’s read post The Ordinal Number of a Term in an Arithmetic Sequence.

RELATED POSTs