nth Term of Linear Sequences

An arithmetic sequence (also called a linear sequence) is a sequence where the difference between consecutive terms is always the same. This fixed difference is called the common difference, $d$.

Examples:

Sequence	Common difference $d$	Pattern
3, 7, 11, 15, 19, …	$+4$	Adding 4 each time
20, 17, 14, 11, 8, …	$-3$	Subtracting 3 each time
1, 1.5, 2, 2.5, 3, …	$+0.5$	Adding 0.5 each time

To find $d$: subtract any term from the next term. $d=T_{n+1}-T_n$.

A sequence is arithmetic if and only if $d$ is constant throughout. If the differences between terms are not all the same, the sequence is not arithmetic (it may be quadratic or geometric — different rules apply).

The term-to-term rule for an arithmetic sequence is simply "add $d$ each time". The nth term formula is more powerful — it lets you find any term directly without listing all the ones before it.

For an arithmetic sequence with first term $a$ and common difference $d$, the nth term is:

$T_n=dn+(a-d)$

This is more commonly derived and remembered as:

$T_n=a+(n-1)d$

Both forms are equivalent. The first form ($dn+c$ where $c=a-d$) makes it easy to read off $d$ as the coefficient of $n$.

What the formula tells you: $T_n$ gives the value of the term in position $n$. $T_1$ is the first term, $T_{10}$ is the tenth term, and so on.

Structure of the formula:

• The coefficient of $n$ is always the common difference $d$.
• The constant is found by substituting $n=1$ (or calculated as $a-d$).

A linear (arithmetic) sequence always produces a formula of the form $T_n=dn+c$ — a straight-line rule in $n$.

Step 1 — Find the common difference $d$ by subtracting consecutive terms.

Step 2 — Write the formula as $T_n=dn+c$.

Step 3 — Substitute $n=1$ and the first term to find $c$: $c=T_1-d$.

Step 4 — Check by substituting $n=2$ (and $n=3$ if unsure).

Worked example 1 — Find the nth term of: 3, 7, 11, 15, 19, …

$d=7-3=4$, so the formula starts $T_n=4n+c$.

Substitute $n=1$: $3=4(1)+c\Rightarrow c=-1$.

$T_n=4n-1$

Check: $T_2=4(2)-1=7$ ✓ and $T_5=4(5)-1=19$ ✓

Worked example 2 — Find the nth term of: 20, 17, 14, 11, …

$d=17-20=-3$, so $T_n=-3n+c$.

Substitute $n=1$: $20=-3(1)+c\Rightarrow c=23$.

$T_n=-3n+23$

Check: $T_3=-3(3)+23=14$ ✓

Once the formula is found, substitute any value of $n$ to find the term at that position.

Worked example — Using $T_n=4n-1$:

$n$	Calculation	Term
1	$4(1)-1$	3
10	$4(10)-1$	39
50	$4(50)-1$	199
100	$4(100)-1$	399

The formula is also used to test whether a particular value is in the sequence. Set $T_n$ equal to the value and solve for $n$ — if $n$ is a positive integer, the value is in the sequence.

Worked example — Is 85 a term in the sequence with $T_n=4n-1$?

$4n-1=85⟹4n=86⟹n=21.5$

$n=21.5$ is not a positive integer, so 85 is not in the sequence.

Is 83 in the sequence? $4n-1=83\Rightarrow 4n=84\Rightarrow n=21$. Yes — it is the 21st term.

GCSE questions often present sequences through patterns or tables rather than as a list of numbers. The method is the same.

Worked example — A pattern of dots: Row 1 has 5 dots, row 2 has 8 dots, row 3 has 11 dots. How many dots in row 20?

The sequence is 5, 8, 11, … with $d=3$.

$T_n=3n+c$; substitute $n=1$: $5=3+c\Rightarrow c=2$.

$T_n=3n+2$

Row 20: $T_{20}=3(20)+2=62$ dots.

Which row has 200 dots? $3n+2=200\Rightarrow 3n=198\Rightarrow n=66$. Row 66 has 200 dots.

Quadratic sequences (where second differences are constant) require a different method — finding the $n^2$ coefficient — and are Higher tier only.

1. Confusing $d$ with the first term

The coefficient of $n$ in the formula is always $d$, the common difference — not the first term. For the sequence 5, 8, 11, … the formula is $3n+2$, not $5n+\text{something}$.

2. Finding $c$ incorrectly

$c=a-d$ where $a$ is the first term. For 5, 8, 11, …: $c=5-3=2$, giving $T_n=3n+2$. A common error is writing $T_n=3n+5$ by using the first term directly as the constant.

3. Not checking the formula

Always substitute $n=1$ and $n=2$ into your formula and compare with the actual terms. This catches sign errors and incorrect values of $c$ before they propagate.

4. Stopping after writing the formula when asked for a specific term

"Find the 50th term" requires substituting $n=50$ into the formula. A formula alone does not answer this question.

5. Concluding a value is in the sequence without checking $n$ is a positive integer

If solving $T_n=k$ gives $n=7.5$ or $n=-3$, the value is not in the sequence. Both conditions must hold: $n$ must be a whole number greater than zero.