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iGCSE (2021 Edition)

9.02 Arithmetic sequences



A sequence in which each term changes from the last by adding a constant amount is called an arithmetic sequence. We refer to the constant the terms are changing by as the common difference, which will result from subtracting any two successive terms $\left(u_{n+1}-u_n\right)$(un+1un).

The progression $-3,5,13,21,\ldots$3,5,13,21, is an arithmetic progression with a common difference of $8$8. On the other hand, the progression $1,10,100,1000,\ldots$1,10,100,1000, is not arithmetic because the difference between each term is not constant.

We denote the first term by the letter $u_1$u1 and the common difference by the letter $d$d. Note that $t_n$tn or $T_n$Tn may also be used instead of $u_n$un

We can find an explicit formula in terms of $u_1$u1 and $d$d, this is useful for finding the $n$nth term without listing the sequence.

Let's look at a table of a concrete example to see the pattern for the explicit formula. For the sequence $-3,5,13,21,\ldots$3,5,13,21,, we have starting term of $-3$3 and a common difference of $8$8, that is $u_1=-3$u1=3 and $d=8$d=8. A table of the sequence is show below:

$n$n $u_n$un Pattern
$1$1 $-3$3 $-3$3
$2$2 $5$5 $-3+8$3+8
$3$3 $13$13 $-3+2\times8$3+2×8
$4$4 $21$21 $-3+3\times8$3+3×8
$n$n $t_n$tn $-3+(n-1)\times8$3+(n1)×8



The pattern starts to become clear and we could guess that the tenth term becomes $u_{10}=69=-3+9\times8$u10=69=3+9×8  and the one-hundredth term $u_{100}=789=-3+99\times8$u100=789=3+99×8. And following the pattern, the explicit formula for the $n$nth term is $u_n=-3+(n-1)\times8$un=3+(n1)×8.

We could create a similar table for the arithmetic progression with starting value $u_1$u1 and common difference $d$d and we would observe the same pattern. Hence,  generating the explicit rule for any arithmetic sequence is given by:



Forms of arithmetic sequences

For any arithmetic sequence with starting value $u_1$u1 and common difference $d$d, we can express it in the explicit form:  $u_n=u_1+\left(n-1\right)d$un=u1+(n1)d


Worked examples

Example 1

For the sequence  $87,80,73,66...$87,80,73,66..., find and explicit rule for the $n$nth term and hence, find the $30$30th term. 

Think: Check that the sequence is arithmetic, does each term differ from the last by a constant? Then write down the the starting value $u_1$u1 and common difference $d$d and substitute these into the general form: $u_n=a+(n-1)d$un=a+(n1)d

Do: Each term is a decrease from the last by $7$7. So we have an arithmetic sequence with: $u_1=87$u1=87 and $d=-7$d=7. The general formula for this sequence is: $u_n=87+\left(n-1\right)\times\left(-7\right)$un=87+(n1)×(7) or $u_n=87-7(n-1)$un=877(n1).

Hence, the $30$30th term is:

$u_n$un $=$= $87-7\left(n-1\right)$877(n1)

Write down the rule.

$\therefore\ u_{30}$ u30 $=$= $87-7\left(30-1\right)$877(301)

Substitute in $n=30$n=30.

  $=$= $87-7\times29$877×29


  $=$= $-116$116



Example 2

For the sequence $10,14,18,22,26,...$10,14,18,22,26,..., find $n$n if the $n$nth term is $186$186.

Think: Find a general rule for the sequence, substitute in $186$186 for $u_n$un and rearrange for $n$n. Finding $n$n is determining which term position has a value of $186$186.

Do: This is an arithmetic sequence with $u_1=10$u1=10 and common difference $d=4$d=4. Hence, the general rule is: $u_n=10+\left(n-1\right)\times4$un=10+(n1)×4, we can simplify this to $u_n=6+4n$un=6+4n, by expanding brackets and collecting like terms. Substituting $u_n=186$un=186, we get:

$186$186 $=$= $6+4n$6+4n
$\therefore4n$4n $=$= $180$180
$n$n $=$= $45$45

Hence, the $45$45th term in the sequence is $186$186.


Practice questions

Question 1

Determine which of the following sequences is an arithmetic progression.

  1. $3,0,-3,-6,\ldots$3,0,3,6,















  2. What is the common difference of this progression?


Question 2

The $n$nth term in an arithmetic progression is given by the formula $T_n=15+5\left(n-1\right)$Tn=15+5(n1).

  1. Determine $a$a, the first term in the arithmetic progression.

  2. Determine $d$d, the common difference.

  3. Determine $T_9$T9, the $9$9th term in the sequence.

Question 3


In an arithmetic progression where $a$a is the first term, and $d$d is the common difference, $T_7=43$T7=43 and $T_{14}=85$T14=85.

  1. Determine $d$d, the common difference.

  2. Determine $a$a, the first term in the sequence.

  3. State the equation for $T_n$Tn, the $n$nth term in the sequence.

  4. Hence find $T_{25}$T25, the $25$25th term in the sequence.





Recognise arithmetic progressions.


Use the formulae for the nth term and for the sum of the first n terms to solve problems involving arithmetic progressions.

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