Finding the position number of a term and number of terms in Arithmetic Progression


In this article, we look at how you can determine the position number of a term in an Arithmetic Progression (A.P.) or Linear Sequence.


Douglas Tawiah Dwumor
April 23, 2024

finding the position number of term and number of terms in Arithmetic Progression

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  • Given common difference, first term the position number of any term can be obtained if the term is known
  • To find the position number of a term, you make n the subject.

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If a term in a linear sequence is given, we can find the position number (n) of the term. That’s, whether the term is 4th or 28th in the sequence. Note that n represents the position number of any term in the A.P. or linear sequence.

Example 1:

In the A.P., 5, 11, 17,..., what term is 101?

Solution

a=5, d=11-5=6, n=? 

U_n=a + (n-1)d

101 = 5 + (n-1)(6)

101=5 + 6n-6

101 = 6n -1

102 = 6n

n=17

Thus, 101, is the 17th term.

Example 2:

Given the sequence -9, -6 \frac{1}{2}, -3 \frac{1}{2},..., where does the term 18 \frac{1}{2} belong?

Solution

a=-9, d=-7.5-(-9)=2.5=6, n=?

Un=a + (n-1)d

18.5 = -9 + (n-1)(2.5)

18.5 =-9 + 2.5n -2.5

18.5 = -11.5 + 2.5n

30 = 2.5n

n=12

Hence, 18 \frac{1}{2} is the 12th term in the sequence.

Finding number of terms in a sequence

If we can find the position number (n) of any term in an A.P. , then we can find the position number of the last term too in a finite sequence. 

The position number of the last term also gives the total number of terms in a finite sequence. 

That’s, given the general term 

Un= a + (n-1)d, 

if Un=l where l is last term,

then n=number of terms in the sequence 

That’s, 

l=a+ (n-1)d 

where n=position number of last term or total number of terms in the finite sequence.

Example 1:

Find the number of terms in the linear sequence 2,-9,-20,...-141.

Solution

a=2, d=-9-2=-11, l=-141, n=?

l=a + (n-1)d

-141=2 +(n-1)(-11)

 -141= 2 -11n + 11

-141 -13=-11n

      -154=-11n

            n=14

The position number of the last term, -141, is 14. Hence, there are 14 terms in the linear sequence.

Example 2:

How many terms are in the linear sequence 5,8,11,.......122 ?

Solution

a=5, d=8-5=3, l=122, n=?

l=a + (n-1)d

122=2 +(n-1)(3)

 122= 2 +3n -3

122= -1+3n

       123=3n

         n=41 

The position number of the last term, 122, is 40. Hence, there are 40 terms in the linear sequence.


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