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.

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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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