C2.7; E2.7 - The nth term of a linear sequence | Cambridge IGCSE International Mathematics

The sequence $5,8,11,14,\ldots$ adds 3 each time. Can you find its hundredth term without writing the other ninety-nine? C2.7 and E2.7 use position-to-term rules to describe linear sequences, calculate distant terms and decide whether a number occurs.

Constant differences

A constant gap makes a sequence linear. Its position-to-term rule, called an nth term, uses the position $n$ directly. Throughout this lesson, $n=1,2,3,\ldots$ : position 1 is the first term, not position 0.

Linear sequence

A linear sequence has a constant first difference between consecutive terms. Its nth term has the form $dn+c$ , where $d$ is that difference and $c$ is a constant adjustment.

Why does $d$ multiply $n$ ? Increasing the position from $n$ to $n+1$ changes $dn+c$ to $d(n+1)+c=dn+c+d$ . The increase is exactly $d$ ; the constant $c$ shifts all terms equally without changing their gaps.

Worked example 1: a positive first difference

Question. For $5,8,11,14,\ldots$ , find the first difference and the coefficient of $n$ in its linear nth term.

Plan. Subtract each term from the one after it.

8-5=3,\qquad11-8=3,\qquad14-11=3.

Answer. The first difference is 3, so the rule has the form $3n+c$ .

Check. The terms $3n$ alone are $3,6,9,12,\ldots$ , with the required gap 3. An adjustment is needed for the starting value, but not for the gap.

The next sequence decreases. Retain the sign of the difference: subtract in the order “next minus current”.

Worked example 2: a negative first difference

Question. For $14,11,8,5,\ldots$ , find the first difference and the coefficient of $n$ .

Plan. Calculate consecutive changes in the direction the sequence runs.

11-14=-3,\qquad8-11=-3,\qquad5-8=-3.

Answer. The difference is $-3$ , so the rule has the form $-3n+c$ .

Check. Increasing $n$ by 1 decreases $-3n+c$ by 3, matching all the given transitions.

A rule $n+3$ adds 3 to the position and has gap 1. It does not describe “add 3 each time”. Term-to-term and position-to-term rules are different instructions.

Key idea

The constant difference gives the coefficient of $n$ , not the constant added at the end of the rule.

Use the sign as well as the size of the gap.

Finding the rule

Once the coefficient is known, match the starting value. For the opening sequence, compare its terms with $3n$ : every term is 2 larger, so the nth term is $3n+2$ .

Position $n$	1	2	3	4
$3n$	3	6	9	12
Required terms	5	8	11	14
Adjustment	$+2$	$+2$	$+2$	$+2$

Linear nth term

For first term $a$ and constant difference $d$ , the nth term is $a+(n-1)d$ . Equivalently, it is $dn+(a-d)$ .

There are $n-1$ steps from the first term to the nth term. This explains the formula and why using $a+nd$ moves one step too far. You can either use it or compare directly with the multiples of $d$ .

Worked example 3: finding a positive rule

Question. Find the nth term of $7,11,15,19,\ldots$ .

Plan. Use the gap 4, then compare the first term with $4\times1$ .

\begin{aligned} d&=11-7=4\\ c&=7-4=3 &&\text{adjust }4n\text{ to reach the first term}\\ \text{nth term}&=4n+3 \end{aligned}

Answer. $4n+3$ .

Check. At $n=1,2,3,4$ , the rule gives $7,11,15,19$ , every supplied term.

The next example keeps the first term but changes the direction. The constant adjustment must be recalculated using the signed coefficient.

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C2.7; E2.7 - The nth term of a linear sequence

Constant differences

Finding the rule

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